Does Light Curve

[Quantoman]

this is a question from another forum that i dont feel

comfortable answering,

 

(only half way through a brian greene book)

 

so can anyone explain simply if it does or does not?

 

ty.

[ajb]

If you mean in a gravitation field then yes.

 

According to general relativity, gravity is the bending of space-time. Light rays now do not move in straight lines but instead follow the anologue of a straight line known as a (null) geodesic.

[swansont]

If you mean in a gravitation field then yes.

 

According to general relativity, gravity is the bending of space-time. Light rays now do not move in straight lines but instead follow the anologue of a straight line known as a (null) geodesic.

 

But within that frame the light is travelling in a straight line.

[ajb]

True, you can set up coordinates that mean that locally the light rays are travelling in straight lines.

 

 

Light rays can be though of as "mapping out" the curvature of space-time. This leads to the prediction that light is bent around a massive object. This was observed by Eddingtion in 1919. This effect is now known as gravitational lensing and is being used to try to detect dark matter, cosmic strings and other objects.

[timo]

In case the question was meant the other way round: Light not also follows the curvature of space, it also contributes to spacetime-curvature.

[ajb]

In case the question was meant the other way round: Light not also follows the curvature of space, it also contributes to spacetime-curvature.

 

 

I was not sure exactly what the question was. Of course Atheist you are right. As light carries energy-momentum it will curve space-time.

[Severian]

But within that frame the light is travelling in a straight line.

 

I had actually typed a response saying exactly this, but then didn't post it because it seemed a little anally retentive.

[[Tycho?]]

Simple answer: from our point of view, yes, light does curve.

 

Complex answer: physics geometry spacetime non-euclidian blah blah blah.

 

If you didn't know that light could be bent, I assume you'll find the simple answer the most useful.

[timo]

... especially since the complex answer seems pretty non-saying and is just a collection of keywords. Feel free to elaborate on that, I am sure some people might be interested in an answer beyond "yes".

[Spyman]

In case the question was meant the other way round: Light not also follows the curvature of space, it also contributes to spacetime-curvature.

I was not sure exactly what the question was. Of course Atheist you are right. As light carries energy-momentum it will curve space-time.

Just to verify if I understand that part correctly:

 

Two parallel laser beams, in the same direction, will not gravitationally attract each other.

(Rest mass for photons = zero.)

 

But if they are in opposite directions and passing each other, then they will both bend.

(Very slightly towards eatch other.)

[woelen]

As light carries energy-momentum it will curve space-time.

Ajb, could you please explain this in more detail? I don't see how light curves space. Of course, light represents a certain amount of energy, propagating through space, but it has no mass. But is it true that energy also contributes to space-time curvature, due to m = E/c²? Or is there another reason for space-time curvature?

[ajb]

Woelen,

 

you can think of it that way.

[[Tycho?]]

Ajb, could you please explain this in more detail? I don't see how light curves space. Of course, light represents a certain amount of energy, propagating through space, but it has no mass. But is it true that energy also contributes to space-time curvature, due to m = E/c²? Or is there another reason for space-time curvature?

 

Energy causes it, no other reason I'm aware of.

[Royston]

Ajb, could you please explain this in more detail? I don't see how light curves space. Of course, light represents a certain amount of energy, propagating through space, but it has no mass. But is it true that energy also contributes to space-time curvature, due to m = E/c²? Or is there another reason for space-time curvature?

 

Light has no 'rest' mass, which isn't the same as zero mass. This is one of the things I queried in my course a couple of months ago.

[woelen]

I'm not an expert at all in this area, but if I understand it well from Snail's post, every object has a mass equal to

 

Mt = M0 + E/c², where M0 is the "rest mass", which we in real life simply call "mass" and E/c² is the relativistic mass? And is it Mt, which determines curvature of space-time, instead of M0? In practical/real life of course, Mt is very close to M0, but fundamentally there are differences.

 

And then, what forms of energy E contribute to this? Kinetic energy seems different to me, because we also have

 

Mt = M0 / √(1 - v²/c²) from special relativity.

 

A series expansion, first order in v²/c² gives:

 

Mt ≈ M0 * (1 + ½v²/c²) + O((v²/c²)²)

 

This gives Mt ≈ M0 + Ekin/c² when |v| is much smaller than c.

 

This is close to my initial guess, but not the same . There is some rest term, depending on the fourth power of v/c. And how are things going for massless things?

 

This is really surprising to me. And now I'm doing some math on it, it strikes me . If one of the physics experts could point me in some direction (a good site), or even better, elaborate more on this over here, that would be very nice.

[swansont]

Light has no 'rest' mass, which isn't the same as zero mass. This is one of the things I queried in my course a couple of months ago.

 

 

Yes, you can define relativistic mass. But now that you have more than one definition for "mass," you need to ascertain the relevant term that goes into the stress-energy tensor.

[Royston]

Yes, you can define relativistic mass. But now that you have more than one definition for "mass," you need to ascertain the relevant term that goes into the stress-energy tensor.

 

Field equations are beyond my scope for now, I'm doing some background reading before I tackle the maths...the next couple of years of courses covers this, I'm happy I understand the definitions and implications at my level, it's all that's needed for the time being.

[timo]

And now I'm doing some math on it, it strikes me . If one of the physics experts could point me in some direction (a good site), or even better, elaborate more on this over here, that would be very nice.

I´d really like to elaborate on the topic but I had to look up too many things for a coherent response and I absolutely don´t have the time for that right now - poke me again in two weeks if there´s questions left over, I´m in principle interested in getting a bit better understanding of the stuff myself. So in short: The directions are: Field equations of GR (= Einstein equations), Energy-momentum tensor (also called Energy-momentum-stress tensor) and relativistic hydrodynamics.

 

I am no fan of relativistic mass myself. I don´t see how it is any good and especially in GR it seems a pretty useless concept. Relativistic mass tends to be used by people in the context of "gravitation is caused by relativistic mass" which seems a questionable statement to me.

 

Let´s take a look of what I remember off my head:

The source of electromagnetism is the electrical current which, in relativistic notation, is a four-vector (in the following just called vector, add it to your list of things to look up if you don´t know that concept). For an electric charge density C and no electrical current it´s simply (C,0,0,0) up to a constant factor. For a charge density C and a current [math] \vec j [/math] it is something like (C,jx,jy,jz). In a suitable gauge the field equations for the electromagnetic field are [math] \square A = j [/math] where [math]\square[/math] is the D´alembert operator (a 2nd derivative) and A is the four-potential that describes the field (if you want the E-field and the B-field used in classical electrodynamics you can take suitable derivatives of A to obtain them). Electromagnetism has the nice property that it is compatible with SR which loosely speaking is because that notation uses four-vectors.

 

The natural attempt for a gravitational theory compatible with SR therefore is to try the same trick: Take the matter-density and the matter current (whatever that shall be) and try to find a vector potential A describing the gravitational field For reasons I cannot remember out of my head this does not work - the reason was that some objects have a different behavior under coordiante transformation that that what would be required from a four-vector field equation.

It turns out that the field equations for GR are tensor equations of rank two which, for example can be written in terms of a matrix. The question is: How is the matter which should be the cause the source of the gravitational field described in terms of such an object? It is described by the energy-momentum tensor. Sadly, I cannot tell you too much about that object without looking it up but let´s take some simplified version which might already show why I don´t like the "gravitation is caused by relativistic mass" statement:

 

For an energy density D, no movement of the masses and vanishing pressure (pressure can add significant terms, especially for photons), the energy-momentum tensor is [math]\left( \begin{array}{cccc} D & & & \\ & 0 & & \\ & & 0 & \\ & & & 0 \end{array}\right) [/math] where I have omitted any constants that might appear (I always omit constants). From that I called D an energy-density you might think "so it is relativistic mass that gravitation couples to". Well, imho that´s not really true. I have assumed "no movement, no pressure" so it´s just the same as mass-density. So let´s see what happens when you go to another coordinate system (the trick about relativistic field equation is that they hold true in any coordinate system!). I am too lazy to really calculate it and I already typed much more that I originally intended to, so just an heuristic argument: energy-density is energy divided by volume. Applying a Lorentz transformation on that, the energy gets a factor gamma. But: Due to what one usually calls length contraction, the volume gets a factor 1/gamma at the same time. Therefore, D -> gamma² D (thinking of it, I think that transformation rule was exactly the reason why a gravitational field equation must be tensorial and not vectorial as in electrodynamics). That´s not really what I´d expect from a coupling to relativistic mass which only gets a factor gamma^1 under transformations.

There´s two catches here:

1) I have completely omitted the other components of the energy-momentum tensor.

2) I have not even said what gravity is supposed to be . In fact, I don´t really know it myself. I tend to see gravitation as a distortion from flat spacetime but that really seems just a non-saying simplification once you start giving the statement a few second thoughts.

 

Ok, time to finish that post, a short summary:

- The important keywords you might want to look up are mentioned above.

- Internal energies of a system (heat and the pressure can play a significant role) also contribute to the field equations. However, they are macroscopic properties (thermodynamics). A microscpopic treatment (gravitational field caused by a single elementary particle) might get along without those.

- I cannot prove that the statement "gravitation is caused by relativistic mass" is wrong, because I don´t know how the microscopic treatment of GR looks like and I therefore don´t even understand the statement. But the transformation properties of the energy-momentum tensor make me think that the statement is wrong.

 

@Spyman: One might think it is correct. For the two parallel beams you could think of asymtotically transform into a system in which the energy of the beams vanishes if they are parellel while there´s no such transformation if they are not parallel. But I am not convinced if that idea would really stand a rigorous investigation.

 

Additions, elaborations and corrections are highly welcome!

[ajb]

I would avoid the term relativistic mass at all costs. Mass is a parameter in the Lagrangian that would correspond to the so called rest mass of a particle.

[woelen]

Atheist, thanks very much for your lengthy post. I learn a lot of it. I also perfectly understand that such materials cannot be covered in detail in a single post and using the keywords you supplied I already found out something.

 

I am familiar with the Maxwell equations of electromagnetism, and had never seen your formulation using the d'Alembert's operator, but it is a very powerful representation. Most striking is that both electrical fields and magnetic fields and their interaction now are included in a single equation, which is easier to memorize than Maxwell's equations. I like easy to memorize equations .

 

I have one minor question:

 

I was thinking that □ = (i/c*∂/∂t, ∂/∂x, ∂/∂y, ∂/∂z) (here i is the imaginary number i*i = -1, c is the speed of light in vacuum).

 

Your □ seems to me to be the inner product □•□ (also written as □²). Its value is

 

-(1/c²)*∂²/∂t² + ∂²/∂x² + ∂²/∂y² + ∂²/∂z²

 

Am I right? With these I can derive Maxwell's equations from the formula □²A=j (and not □A=j).

 

I still have to look into the equations for gravity, they look much more complicated to me due to the tensor algebra, but at least I perfectly understand this formulation for electromagnetism now and have a starting point for gravity.

[timo]

I am familiar with the Maxwell equations of electromagnetism, and had never seen your formulation using the d'Alembert's operator, but it is a very powerful representation. Most striking is that both electrical fields and magnetic fields and their interaction now are included in a single equation, which is easier to memorize than Maxwell's equations. I like easy to memorize equations .

Treating electric and magnetic interaction in one object also makes the fact that it is an electromagnetic interaction nicely visible. The notation I used more or less simply is the relativistic notation (often called covariant notation) of the same. Yes, it looks more elegant to me, too. In fact, most, if not all, relativistic equations look more elegant when written in relativistic notation.

 

I have one minor question:

 

I was thinking that □ = (i/c*∂/∂t, ∂/∂x, ∂/∂y, ∂/∂z) (here i is the imaginary number i*i = -1, c is the speed of light in vacuum).

 

Your □ seems to me to be the inner product □•□ (also written as □²). Its value is

 

-(1/c²)*∂²/∂t² + ∂²/∂x² + ∂²/∂y² + ∂²/∂z²

 

Am I right? With these I can derive Maxwell's equations from the formula □²A=j (and not □A=j).

I know the convention [math] \square = \partial^{\mu} \partial_{\mu} = \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2} [/math] or, in a notation that you sometimes see but that imho is slightly outdated, [math] \square = \partial^{\mu} \partial_{\mu} = - \frac{\partial^2}{\partial t^2} + \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} [/math] (as always, I omit constants i.e. c=1 for me) - it actually depends on the choice for the metric tensor of Minkowsky (flat) spacetime which can be chosen as g=diag(1,-1,-1,-1) or g=diag(-1,1,1,1), respectively.

So yes, my □ would be your □². But I am not sure if I´ve ever seen the notation you are using and I can barely imagine that it´s very usefull for GR because with the factor i on the time-derivative you are actually hiding a very important point of GR, namely that the metric is not euclidian like g=diag(1,1,1,1) (note the signs, strictly speaking it´s not even a metric).

Quick convention suggestion:

 

[math] \partial^{\mu} = \left( \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) [/math]

 

[math] g_{\mu \nu} = diag(1,-1,-1-1) [/math]

 

[math] \partial_{\mu} = g_{\mu \nu} \partial^{\nu} [/math] <- this is in fact the rule to raise and lower indices, not just for the derivative operator; summing over (contracting) two indices should only happen when one is an upper one and the other is a lower one, otherwise there´s probably an error somewhere.

 

[math] \Rightarrow \square = \partial^{\mu} \partial_{\mu} = g^{\mu \nu} \partial_{\mu} \partial_{\nu} = \dots [/math]

 

Oh, and of course you´ll have to add the 1/c in the definition of the partial derivative if you don´t take c=1.

[Spyman]

@Spyman: One might think it is correct. For the two parallel beams you could think of asymtotically transform into a system in which the energy of the beams vanishes if they are parellel while there´s no such transformation if they are not parallel. But I am not convinced if that idea would really stand a rigorous investigation.

PUH ! This discussion has rised to a level far to high for me.

(I tried to follow the answer to woelen - got lost fast.)

 

From my simple point of view there can only be three different outcomes of such an experiment:

 

1) The beams would bend independent of direction. (Always bends.)

 

2) The beams would bend dependent of direction.

 

3) The beams would never bend. (Independent of direction.)

 

If light curves space-time I would think nr. 3 is ruled out.

 

So what you are saying here is that nr. 2 might be correct but implies that you would give nr. 1 a higher propability ?

 

(Or maybe you mean that they bend dependent of direction but if they are parallel there will be some cind of special condition ?)

[woelen]

@Atheist: Give me some time to play around with the tensor stuff. I have no practical experience in using them in this context, but I grasp the idea of the "metric" you gave (indeed, it is not a real metric, IIRC a metric only is a true metric if its eigenvalues are all positive).

 

@Spyman: The answer to your question depends on the nature of the light. If you can regard them as small packets (VERY short pulses of energy, wavelet shape), then I would expect that the light is always bent towards each other, except in the special cases (1) that they are both moving in parallel, in the same direction, (2) that they are on the same line, either moving in the same direction, or moving in opposite direction.

 

Why? Each packet of light curves space, in a similar way as a hole, which kids use with pebbles. If a pebble is moving through the hole, but not through the center, then its path is bent. It follows the shortest path through the hole.

 

But the path of a pebble is not bent, if it goes precisely through the center of a hole. It just moves downward, and upwards again. This explains (2). Situation (1) is somewhat more involved. But imagine that as a hole, which moves at the same speed as the pebble. Effectively, the pebble still can move in a straight line.

 

About situation (2) I am not 100% sure. It might be that in that case, the beams of light do bend towards each other. The reason for that is that the space also is curved in the direction, perpendicular to the path of the beams. But whether this is felt by the light or not depends on how attraction works (rest mass, or total mass?). If total mass determines attraction, then the beams move towards each other, otherwise they follow a straight line. Maybe Atheist can give more info on this.

 

But why not when the beams are moving in

 

I also expect that if the beams are very long (a continuous stream of light for many seconds)

[Spyman]

/Snip/ @Spyman: The answer to your question depends on the nature of the light. If you can regard them as small packets (VERY short pulses of energy, wavelet shape), then I would expect that the light is always bent towards each other, except in the special cases (1) that they are both moving in parallel, in the same direction, (2) that they are on the same line, either moving in the same direction, or moving in opposite direction. /Snip/

We seems to agree, but Atheist is not convinced.

 

/Snip/ Situation (1) is somewhat more involved. But imagine that as a hole, which moves at the same speed as the pebble. Effectively, the pebble still can move in a straight line. /Snip/

The pebbles path would change even if the hole moves along with the same speed.

(Try to put a pepple in the slope of a hole, both does move with the same speed.)

 

Also the center of the hole moves with c, but the edge will move out with the speed of c.

(Exept in the direction, since both move with c, the center is on the front edge.)

 

/Snip/About situation (2) I am not 100% sure. It might be that in that case, the beams of light do bend towards each other. The reason for that is that the space also is curved in the direction, perpendicular to the path of the beams. But whether this is felt by the light or not depends on how attraction works (rest mass, or total mass?). If total mass determines attraction, then the beams move towards each other, otherwise they follow a straight line. /Snip/

If the light already are on the same line then they cant "bend" any closer but the beams should lose energy when in opposite direction.

[woelen]

 

[math] \Rightarrow \square = \partial^{\mu} \partial_{\mu} = g^{\mu \nu} \partial_{\mu} \partial_{\nu} = \dots [/math]

 

Oh' date=' and of course you´ll have to add the 1/c in the definition of the partial derivative if you don´t take c=1.[/quote']

OK, I promised to come back on this. I studied the Einstein representation and tried to evaluate your expression, but I find something peculiar and confusing.

 

Given your definitions, I find the following:

 

[math] \Rightarrow \square = \partial^{\mu} \partial_{\mu} = g_{\mu \nu} \partial^{\mu} \partial^{\nu} = \dots [/math]

 

This happens to have the same value as your expression, because all elements of g have absolute value 1, but for any metric, which has other values than 1 or -1 on its diagional there is a difference. In your expression I end up with third powers of the diagonal elements of g, while I only expect linear expressions in the elements of g.

 

I'm confused ... Apparently I'm missing something? My indexes are upper where yours are lower and vice versa.