Caruthers Posted November 8, 2021 Share Posted November 8, 2021 When constructing a space-time diagram using Minkowski's methodology the time axis is defined as ct. C × t = distance This is akin to saying - how far does light travel when it travels one meter. The value of c is large, therefore to create a meaningful diagram the values of x must be correspondingly large. Perhaps a suitable scale would be light-seconds. In the case of an object traveling at c, then in one second it will travel one light-second. The time axis in a space-time diagram is scaled also in light-seconds (ct for 1 second). Doesn't this simply mean that the time taken to travel one light-second is one second, or in reverse the distance light travels in one second is one light second. Why aren't the axes, for example, t (time in seconds) and x (distance in light-seconds). Link to comment Share on other sites More sharing options...

swansont Posted November 8, 2021 Share Posted November 8, 2021 20 minutes ago, Caruthers said: When constructing a space-time diagram using Minkowski's methodology the time axis is defined as ct. C × t = distance This is akin to saying - how far does light travel when it travels one meter. The value of c is large, therefore to create a meaningful diagram the values of x must be correspondingly large. Perhaps a suitable scale would be light-seconds. You can pick whatever values are suitable. You can pop on whatever exponent of 10 is appropriate, like light-seconds (x10^-6) if your distances are in km But you need the factor of c so that you have consistent units, when doing things like constructing a spacetime interval. Link to comment Share on other sites More sharing options...

studiot Posted November 8, 2021 Share Posted November 8, 2021 Quote Why is the time axis in a space-time diagram a distance Is it a distance ? or is it a composite which has the same mechanical dimensions as distance so is measured in the same units as distance ? If you had mixed axes what would be the units of measurement along an arbitrary line using those axes ? [math]\sqrt {{{\left( {{\rm{distance}}} \right)}^2} - {{\left( {{\rm{time}}} \right)}^2}} [/math] What sort of a unit is this ? Link to comment Share on other sites More sharing options...

Caruthers Posted November 8, 2021 Author Share Posted November 8, 2021 4 hours ago, swansont said: You can pick whatever values are suitable. You can pop on whatever exponent of 10 is appropriate, like light-seconds (x10^-6) if your distances are in km But you need the factor of c so that you have consistent units, when doing things like constructing a spacetime interval. Thank you for your help.. Why can't I pick seconds for the time axis and light-seconds for the distance axis? That way the plot would be velocity. "Consistent units" of plotting distance against distance seems non-intuitive. If we take an object moving at c then at one second it will be at a distance of one light second. This is on a 45 deg line as expected. If an object travels at 0.5c then after 1 second it has traveled 0.5 light seconds. Since ct = x then we must plot ct at 0.5 which is on the same 45 deg line. Any object traveling at any speed seems to lie on the same 45 deg line. 3 hours ago, studiot said: Is it a distance ? or is it a composite which has the same mechanical dimensions as distance so is measured in the same units as distance ? If you had mixed axes what would be the units of measurement along an arbitrary line using those axes ? (distance)2−(time)2−−−−−−−−−−−−−−−−−√ What sort of a unit is this ? I don't quite understand. It only makes sense to plot different units. We do this all the time - distance v time, acceleration v time, height v age, etc. We do not invalidate a plot of height against age because they have different units. Measuring along a curve of height against age is pretty meaningless. A "composite" as you describe it of (say) meters v meters is like saying - how many meters are there per meter. The answer in every case is 1. 1 Link to comment Share on other sites More sharing options...

studiot Posted November 9, 2021 Share Posted November 9, 2021 (edited) 19 minutes ago, Caruthers said: Why can't I pick seconds for the time axis and light-seconds for the distance axis? That way the plot would be velocity. If you plot light-seconds against seconds it is the slope of the plot at any point would be the speed, not the plot itself. This would just be a standard distance v time graph. 19 minutes ago, Caruthers said: I don't quite understand. It only makes sense to plot different units. We do this all the time - distance v time, acceleration v time, height v age, etc. Having different quantities on different axes can make sense but then the plot must be interpreted in a different way. Usually the important quantity is the product of the two quantities on the axes and so the erea under such a graph give the total of the product quantity. For example Work done = Force x Distance so the work carried out by a force moving its point of application a distance is give by the areas under such a graph. You did the right thing in asking the question though, +1, because you need to get it straightened out before tackling Relativity. Edit here is a further example to illustrate the difference. Think about a surveyor measuring the land for a map. He sets up a grid and measures the horizontal distances along his grid to all the points on his grid. Hopefully you can see that both axes on that grid must be in metres. He then has two choices. 1) He can erect vertical distance axis and measure elevations of the points above his base grid, perhaps plotting contours, perhaps picking out building steps , roofs etc. Again this axis must be in metres. 2) He can take a magnetic field meter or a gravimeter, or a lightmeter and record the readings at points on his base grid. He can then use the readings to plot contour graphs of magnetic field strength or gravitational strength or sunshine falling. The Minkowski world has to be like case (1) because it is a measuring framework for the world in which we operate. Case (2) is a different sort of measurement - it is a 2D version of a force - extension graph for a spring and offers different information. Having length on some axes and time on another would make the reference framework like case (2) not like case (1) Edited November 9, 2021 by studiot Link to comment Share on other sites More sharing options...

Caruthers Posted November 9, 2021 Author Share Posted November 9, 2021 Yes, surely it is supposed to be a plot of distance against time. I thought the essence of a space-time diagram is to represent a one dimensional space against time. If that is correct then any point plotted does represent a velocity. I am not sure I understand the analogy to the surveyor.. The surveyor plotting 2 or 3 dimensions in space in order to create a map is not considering time. If the diagram plots distances against distances where does time appear? and how can we call it space-time? Link to comment Share on other sites More sharing options...

Eise Posted November 9, 2021 Share Posted November 9, 2021 What Studiot is aiming at, is that a space-time diagram is not a plot, but a map. However, you cannot put time into a drawn map. So it makes sense to use time multiplied by a velocity to get a distance. 'c' is used because we already know from relativity the importance of 'space-time distance', which is similar to the normal 3-D distance in space, but not the same: instead of the 'space version' of Pythagoras (s^{2} = x^{2} + y^{2} + z^{2} ) we have s^{2} = x^{2} + y^{2} + z^{2} - (ct)^{2}, or in just one space dimension s^{2} = x^{2} - (ct)^{2}, which means that distances are distorted compared to a normal map of Euclidian space. Still, it can be use to create a mechanical device that transforms distances in space.time correctly (see e.g. the space-time globe). 2 Link to comment Share on other sites More sharing options...

studiot Posted November 9, 2021 Share Posted November 9, 2021 11 hours ago, Caruthers said: I am not sure I understand the analogy to the surveyor.. Once again you are right to question that which you do not understand, Eise has caught my drift. 5 hours ago, Eise said: What Studiot is aiming at, is that a space-time diagram is not a plot, but a map. However I am more concerned with this problem that needs to be resolved first of all. 11 hours ago, Caruthers said: Yes, surely it is supposed to be a plot of distance against time. I thought the essence of a space-time diagram is to represent a one dimensional space against time. If that is correct then any point plotted does represent a velocity. This is very important because you have a basic misunderstanding here. Here is a sketch of a distance-time graph linked to a speed-time graph. They have common sections A_B and B_C In the section A_B the body is proceeding at constant speed (note I say speed not velocity since we are only plotting the magnitude, not the direction), So the distance is increasing linearly form A to B. At B the body stops still so time continues but the distance remains constant from B to C. The lower curve shows the speed from A to B as some (positive) value. Whilst the speed from B to C is also constant at zero. If you measure the slope of the distance - time line you will find it gives you the numerical value of the constant speed. As you can see we are trying to help you. In turn it would help us to have an idea where you are coming from. That is what is your approximate level of knowledge, for instance do you know any calculus ? Link to comment Share on other sites More sharing options...

Caruthers Posted November 10, 2021 Author Share Posted November 10, 2021 Yes, I do recognize that you are trying to help me, and I thank you for that. I am a retired engineer with, finally, a little time to study subjects like this solely for pleasure. I am ok at calculus, and have read through several papers on relativity, but at this moment I am focused on understanding the progression from Einstein's work to the graphical representation. You are also correct, my last post contained a silly error. Speed/velocity is not represented by a point and I do understand that. I guess my fingers were ahead of my brain. Thank you for pointing that out. Coming back to the topic, just to be sure we agree on terminology, a map is a graphical representation of the location (e.g. latitude, longitude, elevation) of various features. I understand your analogy of a space-time diagram to a map as both axes are distances, however that brings me back to my original question - if it is a "map" of distances how is time represented? If we consider our diagram to be scaled in units of, say, million meters/second then an object travelling a c will reach 300 (approx) Mm/s after 1 second. Now we need to decide where it will be on the y axis. Given that the y axis is ct then we can say it is at c X 1 second = 300 and we can locate that point. Now lets consider an object travelling a 0.5 c. After 1 second it will be at 150 M m/s. As before we determine the value on the y axis which would again be 300 X 1 second. Now we have an object "mapped" to 300 on one axis and 150 on the other and we can plot that, however we have no representation of time and cannot determine the value of time at that point simply be looking at the diagram. The diagram is representing space-space not space-time. I hope I have explained this properly and that my terminology works for you, and please remember this is a subject where I am in the learning phase. Link to comment Share on other sites More sharing options...

studiot Posted November 10, 2021 Share Posted November 10, 2021 9 hours ago, Caruthers said: Yes, I do recognize that you are trying to help me, and I thank you for that. I am a retired engineer with, finally, a little time to study subjects like this solely for pleasure. I am ok at calculus, and have read through several papers on relativity, but at this moment I am focused on understanding the progression from Einstein's work to the graphical representation. You are also correct, my last post contained a silly error. Speed/velocity is not represented by a point and I do understand that. I guess my fingers were ahead of my brain. Thank you for pointing that out. Coming back to the topic, just to be sure we agree on terminology, a map is a graphical representation of the location (e.g. latitude, longitude, elevation) of various features. I understand your analogy of a space-time diagram to a map as both axes are distances, however that brings me back to my original question - if it is a "map" of distances how is time represented? If we consider our diagram to be scaled in units of, say, million meters/second then an object travelling a c will reach 300 (approx) Mm/s after 1 second. Now we need to decide where it will be on the y axis. Given that the y axis is ct then we can say it is at c X 1 second = 300 and we can locate that point. Now lets consider an object travelling a 0.5 c. After 1 second it will be at 150 M m/s. As before we determine the value on the y axis which would again be 300 X 1 second. Now we have an object "mapped" to 300 on one axis and 150 on the other and we can plot that, however we have no representation of time and cannot determine the value of time at that point simply be looking at the diagram. The diagram is representing space-space not space-time. I hope I have explained this properly and that my terminology works for you, and please remember this is a subject where I am in the learning phase. Yeay excellent progress! A retired Engineer is good, some calculus is better, but not not essential now we have agreed that speed is given by dx/dt. If you would like to indicate what sort of Engineer, (electrical, mechanical etc) I will try to pick my examples from that. The important point is that the axes (x,y,z,t or x,y,z,ct) form a euclidian framework to work with. Einstein originally worked with the first set but his friend and maths advisor, Minkowski, realised that the second set offered theoretical and calculational simplicity, but sadly died not long afterwards. Minkowski took the standard euclidian distance between two points A (x1 , y1, z1) and B (x2, y2, z2, ) which is given by [math]\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} [/math] and extended it to include time. So he faced the same question you are asking which is the time intervals are not lengths. Einstein had already implicitly covered this in his rendering of his Special Relativity equations, including the multiplication by c, but Mikkowski's is more elegant. Minkowski offered that if we not only convert the fourth axis of the framework to a length but get it to point somewhere other than along an already used space axis, we can extend the above formula for the 'distance' between two points, making the time axis a length but a bit different from the others. He did this by multiplying by ic so the expression becomes [math]\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2} + {{\left( {ic{t_2} - ic{t_1}} \right)}^2}} [/math] or [math]\sqrt {\Delta {x^2} + \Delta {y^2} + \Delta {z^2} + \Delta {{\left( {ict} \right)}^2}} = \sqrt {\Delta {x^2} + \Delta {y^2} + \Delta {z^2} - \Delta {{\left( {ct} \right)}^2}} [/math] This was a masterstroke. I will explain further and draw some diagrams once you have confirmed you are OK with this maths. Link to comment Share on other sites More sharing options...

Caruthers Posted November 10, 2021 Author Share Posted November 10, 2021 Does i denote an imaginary number? Link to comment Share on other sites More sharing options...

MigL Posted November 10, 2021 Share Posted November 10, 2021 (edited) I'm not sure I understand what the problem is. On a regular graph, is there a requirement that X and Y must share the same units ? Or can they simply be related by a constant of proportionality; in this case ... c ? Edited November 10, 2021 by MigL Link to comment Share on other sites More sharing options...

Caruthers Posted November 11, 2021 Author Share Posted November 11, 2021 12 hours ago, MigL said: I'm not sure I understand what the problem is. On a regular graph, is there a requirement that X and Y must share the same units ? Or can they simply be related by a constant of proportionality; in this case ... c ? Not really a problem, I am just asking about something I don't, yet, understand. Yes, you can plot anything against anything and some times it is meaningful, and sometimes it isn't. In this case I am learning about space-time diagrams. I had assumed that one of the axes had to be time. I understand that in order to simplify the diagram one dimension of space is often shown. In that case I expected the other would be time. Apparently that is not the case as, in order to create the equations, all axes must be in the same units and are expressed as distances. The x axis is the distance an object moves and the y axis is the value of ct, which is also a distance. I suppose that the value of t is the time that the object takes to reach the distance x, however multiplying by c, which has units of distance/time, results in a distance. This seems to me to result in determining - what is the distance an object travels when it travels a distance x, which obviously doesn't make sense, so I am asking what is the right way to understand the use of ct as an axis. My question then is how is time represented in a space-time diagram. When you say "related by a constant of proportionality" taking a simple case such as a constant of, say, 300 then should we say something like y (meters) = 300x (meters) ? Taking an object moving along the x axis 300 units in 2 seconds then the value of y (ct) would be 600, i.e the point plotted would be at 300 units in one direction and 600 in the other. This has been explained to me as not being a "regular graph" but should be thought of as a map. I guess my basic question is what is the meaning of this type of diagram and where is time represented in a space-time diagram. Link to comment Share on other sites More sharing options...

MigL Posted November 11, 2021 Share Posted November 11, 2021 The term (ic) has to be appended to the t term for calculating the interval length, which has to be a 'distance' in space-time. It does not need to be appended to a space-time diagram, as it will not change the shape of any 'curves' plotted in your space-time diagram. Try it with X and Y on graph paper, then substitute 2X for every X value. The only thing affcted is the spread along the X axis. The difference on a space-time diagram would be that the diagonal ( representing the boundry between spacelike and timelike motion ) has a slope of 1 instead of c . 1 Link to comment Share on other sites More sharing options...

Caruthers Posted November 11, 2021 Author Share Posted November 11, 2021 18 hours ago, studiot said: Yeay excellent progress! A retired Engineer is good, some calculus is better, but not not essential now we have agreed that speed is given by dx/dt. If you would like to indicate what sort of Engineer, (electrical, mechanical etc) I will try to pick my examples from that. The important point is that the axes (x,y,z,t or x,y,z,ct) form a euclidian framework to work with. Einstein originally worked with the first set but his friend and maths advisor, Minkowski, realised that the second set offered theoretical and calculational simplicity, but sadly died not long afterwards. Minkowski took the standard euclidian distance between two points A (x1 , y1, z1) and B (x2, y2, z2, ) which is given by (x2−x1)2+(y2−y1)2+(z2−z1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√ and extended it to include time. So he faced the same question you are asking which is the time intervals are not lengths. Einstein had already implicitly covered this in his rendering of his Special Relativity equations, including the multiplication by c, but Mikkowski's is more elegant. Minkowski offered that if we not only convert the fourth axis of the framework to a length but get it to point somewhere other than along an already used space axis, we can extend the above formula for the 'distance' between two points, making the time axis a length but a bit different from the others. He did this by multiplying by ic so the expression becomes (x2−x1)2+(y2−y1)2+(z2−z1)2+(ict2−ict1)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√ or Δx2+Δy2+Δz2+Δ(ict)2−−−−−−−−−−−−−−−−−−−−−−√=Δx2+Δy2+Δz2−Δ(ct)2−−−−−−−−−−−−−−−−−−−−−−√ This was a masterstroke. I will explain further and draw some diagrams once you have confirmed you are OK with this maths. As a mechanical Engineer I have worked in aerospace, oil & gas and software simulation.. Should the right hand side of the equation also contain i ? Evaluating the equation assuming 1. The object is traveling only in the x direction and y and z = 0 2. Using units of million meters per second 3. The object travels 300 units in one second (i.e. approximately the speed of light, c) Then we get either Root(300^2 + i.300^2) = root(300^2 - i.300^2) and I don't understand what i is or where it came from ,or if the i is not present in rhe right hand we get Root(300^2 + i.300^2) = root(300^2 - 300^2) and I still don't understand i I am confused 😢 Link to comment Share on other sites More sharing options...

MigL Posted November 11, 2021 Share Posted November 11, 2021 i^{2} = -1 Link to comment Share on other sites More sharing options...

studiot Posted November 11, 2021 Share Posted November 11, 2021 (edited) On 11/10/2021 at 10:06 PM, Caruthers said: Does i denote an imaginary number? Yes, is is the famous square root of minus 1 [math]i = \sqrt { - 1} [/math] 5 hours ago, Caruthers said: Should the right hand side of the equation also contain i ? The purpose of i is twofold. To rotate the axis in an imaginary direction. To turn a + into a minus, so not i does not appear on both sides of the equation since [math]{(ict)^2} = {i^2}{c^2}{t^2} = - 1{c^2}{t^2} = - {c^2}{t^2}[/math] 5 hours ago, Caruthers said: 1. The object is traveling only in the x direction and y and z = 0 Yes I am going to take time out of explaining graphs and frameworks and also because sometimes a peek ahead is motivational and I propose to use the just the x and t axes for this peek. I was going to say that If you are an electrical engineer, that I hope you know enough mechanics to work from Newton's laws to the Principle of Relativity. However you have answered that so I will introduce the diagrams. Quote Roy Turner University of Sussex dept of Theoretical Physics. Physics is about measurement and the relationship of one measurement to another. Nowhere is this more true than in Relativity. It is worth knowing that there are lots of different ways to arrive at SR and GR, but they all start with two basic Principles for SR and a third is added for GR. Different authors express these Principles in different ways, most suitable to their route. The strategy I am following is to look for relationships where one measurement is identical to another. Such situations lead to invariants. One such is called the form invariant, which is employed in the Principle of Relativity The idea is that a physical law must not depend upon the coordinate system. That is it should have the same form in all relevant coordinate systems. As an example take two particles A and B, interacting though some force F(x_{A}, x_{B}) and consider their equations of motion in one dimension (the x axis) Newtons tells us that if m_{A} and m_{B} are their respective masses and x_{A}, x_{B} their x coordinates then [math]{m_A}\frac{{{d^2}x}}{{d{t^2}}} = F\left( {{x_A},{x_B}} \right)[/math] and [math]{m_B}\frac{{{d^2}x}}{{d{t^2}}} = - F\left( {{x_A},{x_B}} \right)[/math] Now suppose we describe these same two particles in acoordinate system whose origin is at x_{0} in the original system and call this new system the x' system. Then x_{A} = x'_{A} + x_{0} and x_{B} = x_{B} + x_{0} Substitute these two equations into the first two we get [math]{m_A}\frac{{{d^2}x'}}{{d{t^2}}} = F\left( {x{'_A} + {x_0},x{'_B} + {x_0}} \right)[/math] and [math]{m_B}\frac{{{d^2}x'}}{{d{t^2}}} = - F\left( {x{'_A} + {x_0},x{'_B} + {x_0}} \right)[/math] since x_{0} is a constant [math]\frac{{d{x_0}}}{{dt}} = 0[/math] So the forces of interaction do not have the same form (are not form invariant) as in the original coordiante system. If they did they would be [math]{m_A}\frac{{{d^2}x'}}{{d{t^2}}} = F\left( {x{'_A},x{'_B}} \right)[/math] and [math]{m_B}\frac{{{d^2}x'}}{{d{t^2}}} = - F\left( {x{'_A},x{'_B}} \right)[/math] And the new equations depend upon the origin x_{0} of the new system. So Newton's equatiosn, as they stand do not conform to the Principle of Relativity. However all is not lost. the remedy is to work with coordinate differences, not directly with the coordinates so we write F(x_{A},x_{B}) = f(x_{A} - x_{B}) yielding equations of motion in the new system in the required form [math]{m_A}\frac{{{d^2}x'}}{{d{t^2}}} = f\left( {x{'_A},x{'_B}} \right)[/math] [math]{m_B}\frac{{{d^2}x'}}{{d{t^2}}} = - f\left( {x{'_A},x{'_B}} \right)[/math] This is the motivation for using the pythagorian square root of the sum of the squares of the coordinate distances I mentioned in an earlier post. Edited November 11, 2021 by studiot 2 Link to comment Share on other sites More sharing options...

studiot Posted November 12, 2021 Share Posted November 12, 2021 Ok to continue where I left off. Just to point out that I am offering backgound explanations and insights that you won't find in any one textbook. If you just want a formal development, there are many good presentations on offer. The official title is not 'spacetime' but 'the spacetime continuum'. A continuum is a grid or framework of points where at every point the properties of interest are the same. This expresses the idea of homogeneity, which is also inherent in the Principle of Relativity. For instance take a piece of squared graph paper. It is a continuum. A geometric continuum. A finite continuum, since the paper has edges, but a continuum nonetheless. The graph paper has no 'origin'. We can impose an origin at any point by imposing a pair of axes. And impose scales marked on the paper. This then gives a particular coordinate system to the grid. It also gives us a large number of possible different coordinate systems. I will start with static coordinate systems (which do not need time) before proceeding to moving coordinate systems (which do need time). For completeness I will start with one dimensional (ie one axis only, the x axis) coordinate systems. Coordinate systems are also called 'frames' , and will introduce the motivation for this name. The importance of one dimension is that it gives us a definition of length or distance between any two points A , B in the continuum as in fig (1). We use the rather awkward looking square root of the square of the difference in coordinates, x_{A} and x_{B} because this makes the distance from A to B the same as the distance from B to A and also makes this a positive or zero (ie nonnegative) quantity, which is only zero if A and B are the same point. [math]\sqrt {{{\left( {{x_A} - {x_B}} \right)}^2}} = \sqrt {{{\left( {{x_B} - {x_A}} \right)}^2}} [/math] Note. This conforms to the mathematics definition of a metric or distance function. Metrics in Physics can also be negative and Relativity explores what happens when the metric in Physics is negative. Another mathematical nicety is that the modulus or absolute value |x_{A}-x_{B}| can also be used as a metric, but has the disadvantage that it is not differentiable at zero. Work in progress to be continued. 2 Link to comment Share on other sites More sharing options...

Caruthers Posted November 12, 2021 Author Share Posted November 12, 2021 Got it. Very clear and helpful. If is not in a book it should be! Eagerly awaiting the next part. 1 Link to comment Share on other sites More sharing options...

studiot Posted November 14, 2021 Share Posted November 14, 2021 (edited) On 11/12/2021 at 6:19 PM, Caruthers said: Got it. Very clear and helpful. If is not in a book it should be! Eagerly awaiting the next part. Ok so I have been doodling over the weekend, but sadly did not get as far as I had hoped. So here is the next part. Fig 2 is a repeat of Fig 1 but introducing two coordinate systems. In the red corner we have the red system. x', and in the blue corner we have the blue system x. Note the constant p yields a 'transfomation' equation between the two systems. We can also show that the the distance between A and B have the same value in both systems, following the equations in Fig1. That is it is invariant. I will return to the x' = (x - p) equation in more detail, when we get on to motion. In Fig 3 we strike out in 2 dimensions, with two displacement constants, p and q. Now that we are using the full extent of our graph paper, I have tilred the line AB, shown in green. Hopefully it is easy to see that the length of this line is [math]\sqrt {{{\left( {{x_B} - {x_A}} \right)}^2} + {{\left( {{y_B} - {y_A}} \right)}^2}} = \sqrt {{{\left( {x{'_B} - x{'_A}} \right)}^2} + {{\left( {y{'_B} - y{'_A}} \right)}^2}} [/math] although the coordiante differences are different for the red and blue systems. Displacement of a coordinate system is a 'permitted transformation' that simply move the origin about on the graph paper. The other permitted transformation is a rotation of the soordinate system, as in Fig 4 I have gone explicitly through the geometry to show how once again the length AB or its square AB2 is invariant, regardless of the angle of rotation. The key to this is to note that the length AB is the common hypotenuse to two triangles, ABC in the blue system and ABD in the red system. So for rotation also the length is measured the same in both systems That wraps up the static situation, but what ae we actually doing ? We are drawing lines on a piece of graph paper. All lines on that graph paper have a length measured in, say, cm. That is both axis lines, gridlines and plotted lines. To attach meaning to this look at Fig 5, which is a cooling curve for some substance, and a 'plot' of temperature against time. What units is the distance along the curve between points A and B measured in ? Well like all lines on the graph paper, it is measured in cm But, you say, the axes are in seconds and degrees C. No they are also measured, on the graph, in cm. Bot axes are subject to a scale factor (constant in this case) that introduces the necessary conversion of units. Cm to seconds and cm to degrees, remembering that we can only plot cm on the graph paper. This is where you question about c comes in. c is a velocity and velocity is a (variable or constant) scale factor converting time to length. The next installment will introduce moving coordinate systems, which are the basis of relativity. There are three things we must take forward from this. Firstly that there are invariants that are measured the same in all coordinate systems. Secondly The most important one is a function of coordinate difference, not the coordinates themselves. Thirdly that all we have been considering is Geometry, as in the shape and size of things. Edited November 14, 2021 by studiot 1 Link to comment Share on other sites More sharing options...

studiot Posted November 27, 2021 Share Posted November 27, 2021 (edited) On 11/12/2021 at 6:19 PM, Caruthers said: Got it. Very clear and helpful. If is not in a book it should be! Eagerly awaiting the next part. I a bit suprised there are no more questions, but to press on We have established that Minkowski spacetime is no more than a model or description of the background graph paper we are working on. As such all the axes have to be in the same units to be able to construct vectors in it. But to construct Minkowski spacetime and the motion in it we need two more things. Firstly our model or graph is 4 dimensional, but 'rigid body' motion is one dimensional in that it follows one line, although it may be set in 1,2, 3 or 4 dimensions. Clasically Fig 6 gives us this one dimensional motion and it is often convenient to reduce that motion to take place along the x axis alone, at least as a simplified study. Esentially we are extending the static shift of axes, shown in figs 2 and 3, to become an instantaneous shift, constant at any one instant. Thus fig 6 shows that the constant p becomes a variable as time progresses and introduces time as a variable indirectly via the velocity. This leads to the equation x_{A}' = (x_{A} - vt) which we will need to progress on to einstinian relativity. Because we are trying to develop this from geometry, we want to preserve the relationship that the inner (scalar) product of vectors we have to scale the time axis of our four dimensional graph paper by multiplying it by a suitable constant, having suitable units. This is because to be able to form an inner product all axes have to be in the same units, as previously discussed. Einstein's second Postulate for relativity is that this constant is c (Note this is similar to using v in Fig6) However by itself multiplying by c is not enough as we exist in 3 spatial dimensions and we do not want the t axis to point along any of them. That is we want the t axis to point Perpendicular to the line of any one dimension, Perpendicular to the plane of the paper and Perpendicular to the block of 3 D So we introduce i as an operator that performs these functions. i is best defined as in Fig 7 This gives us the necessary features. The scalar or inner product of out coordinates (regardes as vectors) then becomes {s^2} = {x^2} + {y^2} + {z^2} + {\left[ {ict} \right]^2} = {\left( {{x_1}} \right)^2} + {\left( {{x_2}} \right)^2} + {\left( {{x_3}} \right)^2} + {\left( {{x_4}} \right)^2} = {x_\mu }{x_\mu }s2=x2+y2+z2+[ict]2=(x1)2+(x2)2+(x3)2+(x4)2=xμxμ Where I have taken the opportunity to show the beginnings of Einstein's summation convention. Where a suffix appears twice as in (x_submu) (x_submu) it represents a summation of the all the xs and mu is a (dummy) index we 'understand' to run from 1 through 4 (in this case). So {x_\mu }xμ is the vector (4 vector in this case) given by {x_\mu } = \left( {{x_1},{x_2},{x_3},{x_4}} \right)xμ=(x1,x2,x3,x4) This notation is the beginnings of Tensor notation you may see scatterd about a lot in Physics but not used so much in Engineering. It is not necessary to use tensors, you can do just as well using matrices and we are, of course, just using ordinary algebra spiced with a touch of calculus. (If you prefer matrices, as I do, there is a good book Einstein in Matrix Form, Gunter Ludyk, SpringerGraduate Texts in Physics.) This combination of two quantities with different units to form 4 vector in a Minkowski vector space can be extended to kinetic energy and momentum, which when combined by multiplying the energy by i/c gives the full famous equation of Einstein of offered as E = mc^{2} . Returning to a single space dimension and time we are now in a position to derive the Fitzgerald contraction, which was an empirical obervation, from geometry and a bit of algebra, where is becomes a proper mathematical property of the Minkowski spacetime continuum. we call the Lorenz transformation. So if we have a system S' (with coordinates x', ict') moving along the x axis in system S (with coordinates x, ict) with a constant velocity v we have the invariant inner product {s^2} = {x^2} + {\left[ {ict} \right]^2} = x{'^2} + {\left[ {ict'} \right]^2}s2=x2+[ict]2=x′2+[ict′]2 or {s^2} = {x^2} - {c^2}{t^2} = x{'^2} - {c^2}t{'^2}s2=x2−c2t2=x′2−c2t′2 Some derivations make this equation Einstein's second Postulate, but we have derived it from geometry (Einstein derived it in yet another more complicated way) We will also need the equation from fig 6 I said we would need We now introduce the further constraint that the origins in time must match when S' is alongside S as it passes. That is t = t' = o This means that x = vt when x' =0 and x' = -vt when x = 0 Because there is only one relative velocity v, which is common to both systems S and S', apart from the opposite sign for opposite direction. So taking our expression from fig 6 we have x' = \gamma \left( {x - vt} \right)x′=γ(x−vt) and x = \gamma '\left( {x' + vt'} \right)x=γ′(x′+vt′) Where the gammas are yet to be determined expressions. Eliminating x' between the we find t' = \left[ {t - \frac{x}{v}\left( {1 - \frac{1}{{\gamma \gamma '}}} \right)} \right]t′=[t−xv(1−1γγ′)] Putting our expressions for x' and t' in terms of x and t we find {x^2}\left[ {1 - {\gamma ^2} + \frac{{{\gamma ^2}{c^2}}}{{{v^2}}}{{\left( {1 - \frac{1}{{\gamma \gamma }}} \right)}^2}} \right] + 2xt\left[ {{\gamma ^2}v - \frac{{{c^2}{\gamma ^2}}}{v}\left( {1 - \frac{1}{{\gamma \gamma '}}} \right)} \right] + {t^2}\left[ {{c^2}\left( {{\gamma ^2} - 1} \right) - {\gamma ^2}{v^2}} \right] = 0x2[1−γ2+γ2c2v2(1−1γγ)2]+2xt[γ2v−c2γ2v(1−1γγ′)]+t2[c2(γ2−1)−γ2v2]=0 Now this must be true for all values of x and t so is an identity, not just an equation. This means that each of the summed terms must separately be zero, so the coefficients of x^{2}, xt and t^{2} vanish separately. Thus equating the coefficient of t^{2} to zero gives \gamma = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}γ=11−v2c2√ Which hopefully you will recognise as the gamma in the Lorenz transformation. and using the others gives \gamma = \gamma ' Of course there are other possible solutions to the question. So many authors will stress that any solution must reduce from einstinian relativity to Galilean/Newtonian at low relative velocites. We have actually gone the other way building on Newton to reach Einstein. It is pleasing to see Maths and Physics fitting so well together and complementing each other. If you are still with me we can talk about frames and why we call them frames and what we can expect to 'see', measure and calculate in any frame. Edited November 27, 2021 by studiot Link to comment Share on other sites More sharing options...

studiot Posted November 27, 2021 Share Posted November 27, 2021 [math]{s^2} = {x^2} + {y^2} + {z^2} + {\left[ {ict} \right]^2} = {\left( {{x_1}} \right)^2} + {\left( {{x_2}} \right)^2} + {\left( {{x_3}} \right)^2} + {\left( {{x_4}} \right)^2} = {x_\mu }{x_\mu }[/math] 3 hours ago, studiot said: I a bit suprised there are no more questions, but to press on We have established that Minkowski spacetime is no more than a model or description of the background graph paper we are working on. As such all the axes have to be in the same units to be able to construct vectors in it. But to construct Minkowski spacetime and the motion in it we need two more things. Firstly our model or graph is 4 dimensional, but 'rigid body' motion is one dimensional in that it follows one line, although it may be set in 1,2, 3 or 4 dimensions. Clasically Fig 6 gives us this one dimensional motion and it is often convenient to reduce that motion to take place along the x axis alone, at least as a simplified study. Esentially we are extending the static shift of axes, shown in figs 2 and 3, to become an instantaneous shift, constant at any one instant. Thus fig 6 shows that the constant p becomes a variable as time progresses and introduces time as a variable indirectly via the velocity. This leads to the equation x_{A}' = (x_{A} - vt) which we will need to progress on to einstinian relativity. Because we are trying to develop this from geometry, we want to preserve the relationship that the inner (scalar) product of vectors we have to scale the time axis of our four dimensional graph paper by multiplying it by a suitable constant, having suitable units. This is because to be able to form an inner product all axes have to be in the same units, as previously discussed. Einstein's second Postulate for relativity is that this constant is c (Note this is similar to using v in Fig6) However by itself multiplying by c is not enough as we exist in 3 spatial dimensions and we do not want the t axis to point along any of them. That is we want the t axis to point Perpendicular to the line of any one dimension, Perpendicular to the plane of the paper and Perpendicular to the block of 3 D So we introduce i as an operator that performs these functions. i is best defined as in Fig 7 This gives us the necessary features. The scalar or inner product of out coordinates (regardes as vectors) then becomes [math]{s^2} = {x^2} + {y^2} + {z^2} + {\left[ {ict} \right]^2} = {\left( {{x_1}} \right)^2} + {\left( {{x_2}} \right)^2} + {\left( {{x_3}} \right)^2} + {\left( {{x_4}} \right)^2} = {x_\mu }{x_\mu }[/math] Where I have taken the opportunity to show the beginnings of Einstein's summation convention. Where a suffix appears twice as in (x_submu) (x_submu) it represents a summation of the all the xs and mu is a (dummy) index we 'understand' to run from 1 through 4 (in this case). So [math]{x_\mu }[/math] is the vector (4 vector in this case) given by [math]{x_\mu } = \left( {{x_1},{x_2},{x_3},{x_4}} \right)[/math] This notation is the beginnings of Tensor notation you may see scatterd about a lot in Physics but not used so much in Engineering. It is not necessary to use tensors, you can do just as well using matrices and we are, of course, just using ordinary algebra spiced with a touch of calculus. (If you prefer matrices, as I do, there is a good book Einstein in Matrix Form, Gunter Ludyk, SpringerGraduate Texts in Physics.) This combination of two quantities with different units to form 4 vector in a Minkowski vector space can be extended to kinetic energy and momentum, which when combined by multiplying the energy by i/c gives the full famous equation of Einstein of offered as E = mc^{2} . Returning to a single space dimension and time we are now in a position to derive the Fitzgerald contraction, which was an empirical obervation, from geometry and a bit of algebra, where is becomes a proper mathematical property of the Minkowski spacetime continuum. we call the Lorenz transformation. So if we have a system S' (with coordinates x', ict') moving along the x axis in system S (with coordinates x, ict) with a constant velocity v we have the invariant inner product [math]{s^2} = {x^2} + {\left[ {ict} \right]^2} = x{'^2} + {\left[ {ict'} \right]^2}[/math] or [math]{s^2} = {x^2} - {c^2}{t^2} = x{'^2} - {c^2}t{'^2}[/math] Some derivations make this equation Einstein's second Postulate, but we have derived it from geometry (Einstein derived it in yet another more complicated way) We will also need the equation from fig 6 I said we would need We now introduce the further constraint that the origins in time must match when S' is alongside S as it passes. That is t = t' = o This means that x = vt when x' =0 and x' = -vt when x = 0 Because there is only one relative velocity v, which is common to both systems S and S', apart from the opposite sign for opposite direction. So taking our expression from fig 6 we have [math]x' = \gamma \left( {x - vt} \right)[/math] and [math]x = \gamma '\left( {x' + vt'} \right)[/math] Where the gammas are yet to be determined expressions. Eliminating x' between the we find [math]t' = \left[ {t - \frac{x}{v}\left( {1 - \frac{1}{{\gamma \gamma '}}} \right)} \right][/math] Putting our expressions for x' and t' in terms of x and t we find [math]{x^2}\left[ {1 - {\gamma ^2} + \frac{{{\gamma ^2}{c^2}}}{{{v^2}}}{{\left( {1 - \frac{1}{{\gamma \gamma }}} \right)}^2}} \right] + 2xt\left[ {{\gamma ^2}v - \frac{{{c^2}{\gamma ^2}}}{v}\left( {1 - \frac{1}{{\gamma \gamma '}}} \right)} \right] + {t^2}\left[ {{c^2}\left( {{\gamma ^2} - 1} \right) - {\gamma ^2}{v^2}} \right] = 0[/math] Now this must be true for all values of x and t so is an identity, not just an equation. This means that each of the summed terms must separately be zero, so the coefficients of x^{2}, xt and t^{2} vanish separately. Thus equating the coefficient of t^{2} to zero gives [math]\gamma = \frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}[/math] Which hopefully you will recognise as the gamma in the Lorenz transformation. and using the others gives [math]\gamma = \gamma '[/math] Of course there are other possible solutions to the question. So many authors will stress that any solution must reduce from einstinian relativity to Galilean/Newtonian at low relative velocites. We have actually gone the other way building on Newton to reach Einstein. It is pleasing to see Maths and Physics fitting so well together and complementing each other. If you are still with me we can talk about frames and why we call them frames and what we can expect to 'see', measure and calculate in any frame. My apologies something went seriously qwrong with the LATex aftr inputting. Correcting the quote is the only way I can see to put it right. Link to comment Share on other sites More sharing options...

Caruthers Posted December 2, 2021 Author Share Posted December 2, 2021 Before asking additional questions I wanted to see, and hopefully understand, your full explanation and I am still working through bit by bit. I also got sidetracked getting my sailboat ready for winter. As you are aware my focus is on how time is represented. What has become clear is that in these diagrams the slope becomes a factor of c, and that makes sense to me. I am now thinking about what it means to rotate the "time" axis. I get the math, but what does it mean for the direction of time to point in a different direction? Anyway, still studying your excellent presentation of the subject. Link to comment Share on other sites More sharing options...

studiot Posted December 2, 2021 Share Posted December 2, 2021 (edited) 15 minutes ago, Caruthers said: Before asking additional questions I wanted to see, and hopefully understand, your full explanation and I am still working through bit by bit. I also got sidetracked getting my sailboat ready for winter. As you are aware my focus is on how time is represented. What has become clear is that in these diagrams the slope becomes a factor of c, and that makes sense to me. I am now thinking about what it means to rotate the "time" axis. I get the math, but what does it mean for the direction of time to point in a different direction? Anyway, still studying your excellent presentation of the subject. Thank you for this useful information. I fully understand about sailing vessels. My brother goes through this routine twice a year, once to lift the boat in and once to lift it out. The club at Gillingham hires a big crane twice a year to effect this for members. So calendars and timetables have to scramble to meet the appointed dates. So it's no problem to spread your excellent conversation out. As regards rotation, I have been rotating axes for simplicity. The actual spacetime rotation is of what amounts to rotating the position vector of the moving object (a bit) towards the time axis and away from the space axes. This increases the projection of the position vector onto the time axis and decreases it on the space axes - In other words time dilation and length contraction. But to have physical meaning the time axis must be modified by a factor with the units of velocity. I am saying units not dimensions to avoid confusion with the dual use of the word 'dimension' as in MLT etc, and the quantities measured by the coordinate axes. Hopefully you are not in the storm zone with your boat. Edited December 2, 2021 by studiot Link to comment Share on other sites More sharing options...

joigus Posted December 2, 2021 Share Posted December 2, 2021 On 11/10/2021 at 11:57 PM, MigL said: I'm not sure I understand what the problem is. On a regular graph, is there a requirement that X and Y must share the same units ? Or can they simply be related by a constant of proportionality; in this case ... c ? Probably just a rhetorical question, but interesting nonetheless. The answer, very much having to do with the excellent points made, is all about symmetry groups --from a mathematical physics perspective. In economics, eg., you have graphs representing marginal utility --dimensionless, if I remember correctly-- plotted against number of units of goods --again, dimensionless. In electrical engineering tho, you have graphs plotting intensity against voltage --both dimensionful. But the subtle mathematical point is that in neither of those cases is there a group of symmetry relating different, but equaly valid, observational stances --classes of valid observers. In the physics of space-time, there is such class of valid observations, and the measurements of time and space of one observer are mixed up with the measurements of time and space of another, so it is natural to define the parameters of the transformation as dimensionless numbers. You could insist that (I,V) (intensity and voltage) be a 'vector', but it doesn't make much sense unless you can define transformations that take one stance to another and relate them in a linear way: I'=aI+bV V'=cI+dV with a,b,c,d being the dimensionless parameters that mix them up. That's the reason why in mathematical physics you tend to be more careful and say: Such and such quantities are a vector under SO(3) (rotations); or a vector under O(3) (rotations and inversions-reflections), etc. In the case of Minkowski vectors, it's all about inertial observers. It's what we call the 'Lorentz group'. 1 Link to comment Share on other sites More sharing options...

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