univeral theory 5 Posted February 1, 2013 Share Posted February 1, 2013 Dear reader; It seems that iam off target! And I beg your pardon: Withthe help of mathematical illustrations and explanations, may you please show me(step by step) on how to check an equation like E=MC^{2} in dimensional analysis? Please I beg. Link to post Share on other sites

elfmotat 324 Posted February 1, 2013 Share Posted February 1, 2013 Energy is measured in units of Joules. We also know that 1 Joule = 1 kilogram*meter^{2} / second^{2}. Mass is measured in units of kilograms. "c" is a velocity, which is measured in units of meter/second. c^{2} is therefore measured in units of meters^{2 }/ second^{2}. If we multiply mass by c^{2}, we get units of (kilograms)*(meters^{2 }/ second^{2}) = Joules, so we know that the equation is dimensionally correct because mc^{2} has units of energy. Link to post Share on other sites

ajb 1567 Posted February 1, 2013 Share Posted February 1, 2013 I must point out that dimensional analysis can tell you if an expression in physics is necessarily wrong, but is not sufficient to tell you that the expression is correct. For example, [math]E = 2 m c^{2}[/math] is dimensionally sound, buy physically flawed. 1 Link to post Share on other sites

Bignose 946 Posted February 1, 2013 Share Posted February 1, 2013 I must point out that dimensional analysis can tell you if an expression in physics is necessarily wrong, but is not sufficient to tell you that the expression is correct. For example, [math]E = 2 m c^{2}[/math] is dimensionally sound, buy physically flawed. A really great point that cannot be emphasized enough. So, I'm going to write the same thing just worded a little differently: A dimensional analysis is only a first check on how potentially useful an equation is. That is to say, to date there has not been a single useful physics equation that isn't dimensionally sound. But, simply being dimensionally sound is not enough to be considered useful -- true usefulness stems from an equation making accurate predictions with the correct units. Link to post Share on other sites

univeral theory 5 Posted February 2, 2013 Author Share Posted February 2, 2013 A really great point that cannot be emphasized enough. So, I'm going to write the same thing just worded a little differently: A dimensional analysis is only a first check on how potentially useful an equation is. That is to say, to date there has not been a single useful physics equation that isn't dimensionally sound. But, simply being dimensionally sound is not enough to be considered useful -- true usefulness stems from an equation making accurate predictions with the correct units. does this mean that every dimensionally correct equation must be dimensionally sound. and not all dimensionally sound equations are dimensionally correct. and thus dimensional analysis is concerned with the soundness of the equation? Link to post Share on other sites

ajb 1567 Posted February 2, 2013 Share Posted February 2, 2013 does this mean that every dimensionally correct equation must be dimensionally sound. and not all dimensionally sound equations are dimensionally correct. and thus dimensional analysis is concerned with the soundness of the equation?What we mean is that from a collection of parameters or constants with and without dimensions one can build equations that are mathematically sound as the dimension on both sides of the equals sign being the same. (If they did not one would have to extend our collection of constants with dimensions to make it sound; this is one way of looking at the fundamental constants of nature.) This does not mean that the equations we construct are mathematically interesting, but more important to this discussion is that such equations may not be physically relevant. They may not be any good at describing nature, although they are dimensionally sound. Link to post Share on other sites

elfmotat 324 Posted February 2, 2013 Share Posted February 2, 2013 does this mean that every dimensionally correct equation must be dimensionally sound. and not all dimensionally sound equations are dimensionally correct. and thus dimensional analysis is concerned with the soundness of the equation? You seem to be making a rather odd distinction between "dimensionally sound" and "dimensionally correct." I'm not sure what the difference is between the two, so perhaps you could elaborate. Link to post Share on other sites

D H 1371 Posted February 2, 2013 Share Posted February 2, 2013 does this mean that every dimensionally correct equation must be dimensionally sound. and not all dimensionally sound equations are dimensionally correct. and thus dimensional analysis is concerned with the soundness of the equation? This doesn't make sense. What it means is that every correct equation must be dimensionally sound. Dimensional analysis is a handy scheme for rejecting expressions such as 1 kilogram + 1 meter as nonsense. However, that an equation or expression is dimensionally sound does not necessarily mean that it is correct. Think of it in terms of the difference between "if" and "only if". Link to post Share on other sites

x(x-y) 10 Posted February 2, 2013 Share Posted February 2, 2013 Also, it's important to note that one generally obtains the correct equation (both dimensionally sound and physically correct) through means of experimentation (i.e. empirically) not simply by using mathematics - of course, the theoretical derivations can help you to achieve the correct equation for some quantity, however at some point you are going to have to do an experiment(s) in order to obtain a meaningful result which is physically correct. This goes for differential equations too, as you may be able to find the general solution of a differential equation through mathematics - however to obtain the complete solution you will need boundary/initial conditions which are obtained through experiment. 1 Link to post Share on other sites

alpha2cen 18 Posted February 3, 2013 Share Posted February 3, 2013 (edited) Also, it's important to note that one generally obtains the correct equation (both dimensionally sound and physically correct) through means of experimentation (i.e. empirically) not simply by using mathematics - of course, the theoretical derivations can help you to achieve the correct equation for some quantity, however at some point you are going to have to do an experiment(s) in order to obtain a meaningful result which is physically correct. This goes for differential equations too, as you may be able to find the general solution of a differential equation through mathematics - however to obtain the complete solution you will need boundary/initial conditions which are obtained through experiment. In the case of solving mathematical equations, using dimensionless forms of the equations is more reliable. Edited February 3, 2013 by alpha2cen Link to post Share on other sites

x(x-y) 10 Posted February 3, 2013 Share Posted February 3, 2013 In the case of solving mathematical equations, using dimensionless forms of the equations is more reliable. Agreed, but my main point was that obtaining the physically correct equation is not done using mathematics - one must experiment to achieve this. Link to post Share on other sites

swansont 7477 Posted February 3, 2013 Share Posted February 3, 2013 Agreed, but my main point was that obtaining the physically correct equation is not done using mathematics - one must experiment to achieve this. And yet Einstein managed to derive length contraction, time dilation and mass-energy equivalence, so "must" is an over-reach. You need experiments to determine arbitrary coefficients, but not all equations are under-constrained. Link to post Share on other sites

x(x-y) 10 Posted February 4, 2013 Share Posted February 4, 2013 And yet Einstein managed to derive length contraction, time dilation and mass-energy equivalence, so "must" is an over-reach. You need experiments to determine arbitrary coefficients, but not all equations are under-constrained. Yes, thought experiments are all very well - but until they can be confirmed by actual experiments, the "results" are largely meaningless (that is not to say that thought experiments are not useful however). Link to post Share on other sites

elfmotat 324 Posted February 4, 2013 Share Posted February 4, 2013 Yes, thought experiments are all very well - but until they can be confirmed by actual experiments, the "results" are largely meaningless (that is not to say that thought experiments are not useful however). But that wasn't your original point. You were saying that you need physical experiments to obtain valid equations, and swansont pointed out that this is true only when they contain tunable parameters. Obviously your equations need to agree with experiment for them to be meaningful, but nobody was arguing about that. Link to post Share on other sites

swansont 7477 Posted February 4, 2013 Share Posted February 4, 2013 Yes, thought experiments are all very well - but until they can be confirmed by actual experiments, the "results" are largely meaningless (that is not to say that thought experiments are not useful however). Yes, this is to some extent true, but peripheral to the point under discussion: physically correct equations can indeed be obtained via theory. It's true — I gave three examples — so let's move on. Link to post Share on other sites

x(x-y) 10 Posted February 4, 2013 Share Posted February 4, 2013 In reply to both comments above: Indeed, I understand that there are a select few cases where physically correct equations can be obtained purely from theoretical analysis and thus perhaps my wording of "must" was too extreme given these cases; however, my overall point (absent the "must" wording) stands. I appreciate the corrections, thank you. Link to post Share on other sites

Amaton 12 Posted February 10, 2013 Share Posted February 10, 2013 (edited) I've never really been exposed to dimensional analysis, but is it really as simple as... [math]\mbox{E}=\mbox{mass}\times\left(\dfrac{\mbox{length}}{\mbox{time}}\right)^2=mc^2[/math] Of course, this is vaguely simplified, but is this the basic idea? Edited February 10, 2013 by Amaton Link to post Share on other sites

ajb 1567 Posted February 10, 2013 Share Posted February 10, 2013 I've never really been exposed to dimensional analysis, but is it really as simple as... [math]\mbox{E}=\mbox{mass}\times\left(\dfrac{\mbox{length}}{\mbox{time}}\right)^2=mc^2[/math] Of course, this is vaguely simplified, but is this the basic idea? That is the basic idea, or better to say that the dimensions of energy are equal to the dimensions of mass times velocity squared. Note that this holds true for the classical kinetic energy [math]KE = \frac{1}{2}m v^{2}[/math]. But this analysis tells us nothing about the factor of one half here. Link to post Share on other sites

Amaton 12 Posted February 10, 2013 Share Posted February 10, 2013 That is the basic idea, or better to say that the dimensions of energy are equal to the dimensions of mass times velocity squared. Note that this holds true for the classical kinetic energy [math]KE = \frac{1}{2}m v^{2}[/math]. But this analysis tells us nothing about the factor of one half here. Ah, thanks. So for example, let's say, [math]E=mc[/math] is not dimensionally correct? Link to post Share on other sites

John Cuthber 3830 Posted February 10, 2013 Share Posted February 10, 2013 Ah, thanks. So for example, let's say, [math]E=mc[/math] is not dimensionally correct? Quite. So you can tell if something is wrong by dimensional analysis, but you can't tell what is right. Also, you can not, for example, tell what particular mass "m" is. For all you can tell from the dimensional analysis, it might be the mass of the experimenter's mother. Link to post Share on other sites

Amaton 12 Posted February 12, 2013 Share Posted February 12, 2013 Quite. So you can tell if something is wrong by dimensional analysis, but you can't tell what is right. Also, you can not, for example, tell what particular mass "m" is. For all you can tell from the dimensional analysis, it might be the mass of the experimenter's mother. Haha, I see. Much thanks. Link to post Share on other sites

univeral theory 5 Posted February 13, 2013 Author Share Posted February 13, 2013 You seem to be making a rather odd distinction between "dimensionally sound" and "dimensionally correct." I'm not sure what the difference is between the two, so perhaps you could elaborate. What I mean is that an equation can be dimensionally sound, but physically flawed. And is there any physical means by which E=MC^{2} is dimensionally sound but physically flawed? Link to post Share on other sites

ajb 1567 Posted February 13, 2013 Share Posted February 13, 2013 What I mean is that an equation can be dimensionally sound, but physically flawed.That is correct.And is there any physical means by which E=MC^{2} is dimensionally sound but physically flawed?As far as we know E=mc^{2} holds true, when applied correctly. In relativistic theory one often does not make much distinction between mass and energy, due to the fact that we have no evidence to suggest doing so is incorrect. Link to post Share on other sites

summy 2 Posted January 19, 2017 Share Posted January 19, 2017 dimension of E=1/2cvsquared -2 Link to post Share on other sites

Sensei 1045 Posted January 20, 2017 Share Posted January 20, 2017 (edited) For example, [math]E = 2 m c^{2}[/math] is dimensionally sound, buy physically flawed. Actually [math]E = 2 m c^{2}[/math] is energy released during annihilation of two matter-antimatter particles with equal m rest-masses.. Edited January 20, 2017 by Sensei Link to post Share on other sites

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