Questions about the relativistic kinetic energy expressions

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I am asking assistance in addressing several questions I have with the relativistic kinetic energy expressions given as {I am sorry for the format of the notations. It was inadvertently distorted.}

KE=mc2 [1/sqrt(1-(v2/c2)) -1] (1)

and its equivalent presented in the form of a series

KE= mc2 +m (v2/2) + (3/8) m (v4/c2) +(5/16)m(v6/c4) (2)

The following inconsistencies are observed

#1
Because the Eq. 2 does not lead to infinity, even with v approaching c, the question arises about the validity of the inference from Eq. 1 that [1] “This expression approaches infinity as the velocity v approaches the velocity of light c. The velocity must therefore always remain less than c, however great may be the energies used to produce the acceleration.”
Please explain the contradiction.

#2
Einstein states [1] that “In accordance with the theory of relativity the kinetic energy of a material point of mass m is no longer given by the well-known expression m(v2/2) but by the expression (1).” Further, by explaining Eq.2, Einstein [1] notes that “The first term mc2 does not contains the velocity, and requires no consideration if we are only dealing with the question as to how the energy of a point-mass depends on the velocity.” To my mind, this statement essentially nullifies any use of the Eq. 1 for the calculation of the kinetic energy of the point mass because both v and c have to be used in this expression to carry out the calculation.
Please explain the contradiction.

REFERENCES:

1. Einstein, A. “Relativity. The special and General Theory.” Barnes & Noble, 2004.
 
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  • #2
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I am a mechanical engineer who is looking to advance himself in the relativist mechanics.
 
  • #3
Orodruin
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KE=mc2 [1/sqrt(1-(v2/c2)) -1] (1)

and its equivalent presented in the form of a series

KE= mc2 +m (v2/2) + (3/8) m (v4/c2) +(5/16)m(v6/c4) (2)
Almost, the series continues with an infinite number of terms, but for velocities that are sufficiently smaller than c, the truncated series is a good approximation. Also, the second expression is an expression for the total energy, not the kinetic energy. You need to remove ##mc^2## (i.e. the rest energy) to get the kinetic energy.
 
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  • #4
Ibix
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KE=mc2 [1/sqrt(1-(v2/c2)) -1] (1)

and its equivalent presented in the form of a series

KE= mc2 +m (v2/2) + (3/8) m (v4/c2) +(5/16)m(v6/c4) (2)
Note that these two equations are inconsistent. The first is the kinetic energy, ##K=(\gamma-1) mc^2##. The second is a Taylor expansion of the total energy ##E=\gamma mc^2##.
Because the Eq. 2 does not lead to infinity, even with v approaching c,
Doesn't it? The truncated series you've shown doesn't, but the infinite series does.
Einstein states [1] that “In accordance with the theory of relativity the kinetic energy of a material point of mass m is no longer given by the well-known expression m(v2/2) but by the expression (1).” Further, by explaining Eq.2, Einstein [1] notes that “The first term mc2 does not contains the velocity, and requires no consideration if we are only dealing with the question as to how the energy of a point-mass depends on the velocity.” To my mind, this statement essentially nullifies any use of the Eq. 1 for the calculation of the kinetic energy of the point mass because both v and c have to be used in this expression to carry out the calculation.
Please explain the contradiction.
There isn't a contradiction. Einstein is just saying that you can ignore the first term when considering how energy changes with velocity since it doesn't change. It's only the later terms that matter. Or you can just use equation 1.
 
  • #5
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Almost, the series continues with an infinite number of terms, but for velocities that are sufficiently smaller than c, the truncated series is a good approximation. Also, the second expression is an expression for the total energy, not the kinetic energy. You need to remove ##mc^2## (i.e. the rest energy) to get the kinetic energy.
Thank you for the response. Yes, my Eq. 2 meant to be an infinite series. I just missed adding signs of + and... . I would dispute that it is for the total energy. My opinion is based on a) It has been given in some sources as an approximation of the kinetic energy; b) on my superficial analysis, and c) It was shown in Einstein [1] as an expression of the kinetic energy. My main point is that per E. explanation, "...mc2 does not contain the velocity, and requires no consideration if we are dealing with... the energy of a point-mass ..." It is clear. However, one cannot ignore this term in his original expression for the kinetic energy with mc2/sqrt(1-v2/c2). So, this is a conundrum.
 
  • #6
Orodruin
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I would dispute that it is for the total energy.
Then you would be wrong. Sorry to put it so bluntly, but that is the case.

I happen to have [1], although in a Swedish translation, since many years ago when I was a teenager and started getting interested in physics. Where it talks about the series expansion, it most certainly is talking about the expansion of the expression
$$
\frac{mc^2}{\sqrt{1-v^2/c^2}} = mc^2 + \frac{mv^2}{2} + \frac{3mv^4}{8c^2} + \ldots,
$$
the total energy, not the expansion of
$$
mc^2\left[\frac{1}{\sqrt{1-v^2/c^2}} - 1\right],
$$
the kinetic energy.

Einstein does call the first expression "kinetic energy", but in modern terms that would be a misnomer.
 
  • #7
Ibix
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I would dispute that it is for the total energy.
You would be wrong to do so. Look up the Taylor series expansion for ##1/\sqrt{1-x}##.
 
  • #8
Orodruin
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I would also note that [1] was written by Einstein in 1916 as a popular account of his theories. Although Einstein does better than most in not dumbing things down to an extreme level, it is still a popular account and not a textbook. Furthermore, it was written a very long time ago and naturally the modern perspective on relativity is completely missing. I would not read it as a means of learning relativity, but rather as a means of understanding a little bit of how Einstein thought about relativity.
 
  • #9
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Thank you very much for your clarifications. Please bear with me. I usually come up with correct results at the end. (There is the famous expression of Churchill that Americans usually do wright things after they did everything wrong:-). However, my main point is still there (regardless which expression we call the total or kinetic energy). Albert clearly says that at the calculation of kinetic energy one should not use the term mc2 because it does not contain velocity. However (as I mentioned before) one cannot avoid it when using the original (not expanded) energy equation. At a present time, I cannot devote myself to delve in the advanced/modern review of the relativity, but I will try to do a calculation to verify if the series equation leads to infinity. At first glance, I thought it shouldn't. I will be back in a couple of days with more questions. I envision that my brief interaction with the topic of relativity may help me in some of my engineering endeavors.
 
  • #10
Orodruin
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However (as I mentioned before) one cannot avoid it when using the original (not expanded) energy equation.
Yes you can, you just subtract it if you want what is called the "kinetic energy" in modern terms. That term itself is referred to as "rest energy", i.e., the energy the object has in an inertial frame where it is at rest.

At first glance, I thought it shouldn't.
It is quite clear that it does, since it is the expansion of an expression that does diverge as ##v \to c##. If it did not diverge, then it would not be the correct series expansion.
 
  • #11
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Why not just derive the Lorentz transformation from the two postulates, then use proper time to get an expression for relativistic momentum, and then
W = ∫(dP/dt) dx ? Seems a lot cleaner, IMHO.
 
  • #12
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Yes you can, you just subtract it if you want what is called the "kinetic energy" in modern terms. That term itself is referred to as "rest energy", i.e., the energy the object has in an inertial frame where it is at rest.

It is quite clear that it does, since it is the expansion of an expression that does diverge as ##v \to c##. If it did not diverge, then it would not be the correct series expansion.
V. Dunaevsky: According to the literature I have, the expression (1-x^2)^-1/2, to which (1-v^2/c^2)^1/2 can be reduced, converges at abs x<1.
 
  • #13
Orodruin
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V. Dunaevsky: According to the literature I have, the expression (1-x^2)^-1/2, to which (1-v^2/c^2)^1/2 can be reduced, converges at abs x<1.
Yes? This has nothing to do with what you quoted, which discussed the limit ##x \to 1##, not the value of the series for ##|x| < 1##. It is perfectly possible for a function that has a finite value for all ##x \neq a## to diverge in the limit ##x \to a##.
 
  • #14
Ibix
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V. Dunaevsky: According to the literature I have, the expression (1-x^2)^-1/2, to which (1-v^2/c^2)^1/2 can be reduced, converges at abs x<1.
You are missing a minus sign in your second exponent but, assuming that's a typo, agreed (Edit: writing maths is a lot easier with LaTeX - there's a tutorial linked below the reply box). The series converges to a finite sum for all ##|x|<1##. This is not what is meant by diverging as ##x\rightarrow 1##. That means that the sum of the series grows without limit as ##x## approaches 1 - which it does.

Physically, this means that the energy has a finite value for all ##|x|<1##. If it didn't, it would be impossible for a massive object to reach some speed less than ##c##. However it diverges (grows without bound) as we approach ##c##, with the result that ##E\rightarrow\infty## as ##v\rightarrow c##. The series sum also grows without bound as ##x## approaches 1.

Once ##x\geq 1##, the series does not converge to a finite value. Since the original function is either undefined or imaginary, that's not surprising.
 
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You are missing a minus sign in your second exponent but, assuming that's a typo, agreed (Edit: writing maths is a lot easier with LaTeX - there's a tutorial linked below the reply box). The series converges to a finite sum for all ##|x|<1##. This is not what is meant by diverging as ##x\rightarrow 1##. That means that the sum of the series grows without limit as ##x## approaches 1 - which it does.

Physically, this means that the energy has a finite value for all ##|x|<1##. If it didn't, it would be impossible for a massive object to reach some speed less than ##c##. However it diverges (grows without bound) as we approach ##c##, with the result that ##E\rightarrow\infty## as ##v\rightarrow c##. The series sum also grows without bound as ##x## approaches 1.

Once ##x\geq 1##, the series does not converge to a finite value. Since the original function is either undefined or imaginary, that's not surprising.
I am sorry. I don't follow your logic. If we agree that the series converges for all abs x<1, how then the sum of a series can grow without limit (as you suggests) when x approaches 1?
Once we said that the series converges to a finite sum for all abs x < 1, than to me it means that no divergence could take place as x approaches 1.
What then to your mind means all abs<1? To me, it means that if one takes x=0.1 (v=0.1 c) or x=0.999 (v=0.999c), there will be a finite sum for each respective series.
 
  • #16
Ibix
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I am sorry. I don't follow your logic. If we agree that the series converges for all abs x<1, how then the sum of a series can grow without limit (as you suggests) when x approaches 1?
Once we said that the series converges to a finite sum for all abs x < 1, than to me it means that no divergence could take place as x approaches 1.
What then to your mind means all abs<1? To me, it means that if one takes x=0.1 (v=0.1 c) or x=0.999 (v=0.999c), there will be a finite sum for each respective series.
The type of convergence you are talking about is the series convergence - as you increase the number of terms, ##n##, in the series you are summing for some fixed ##x##, does the sum tend towards a finite value? Yes it does, as long as ##|x|<1##. This is true, and is an important part of showing that the series remains a valid approximation for the function when ##x## is close to 1, but it is not relevant to the point @Orodruin and I are making.

The point is that for a given ##x##, where ##x<1##, the series converges to a finite value (##1/\sqrt{1-x^2}##, obviously). For ##x'##, where ##x<x'<1##, it converges to a larger value (##1/\sqrt{1-x'^2}##). For ##x''##, where ##x'<x''<1##, it converges to a still larger value (##1/\sqrt{1-x''^2}##). These values do not tend towards a finite value as you continue using larger and larger ##x##. So the value of the infinite series sum grows without bound as ##x## (not the number of terms you sum) increases.

Sum the series to as many terms as you can for x=0.9. Then x=0.99. Then x=0.999. Keep adding as many nines as you like, and you'll find the value continues to rise - it never stops as you add nines, and the sum doesn't tend towards a finite value. The sum rises without bound (Edit: by which I mean that there is no value you cannot reach by picking a large enough ##x##, ##x<1##) as you increase x. It reaches infinity in the limit of infinite nines, which is a long-winded way of writing x=1. This is exactly the same behaviour as ##1/\sqrt{1-x^2}##.
 
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  • #17
Orodruin
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What then to your mind means all abs<1? To me, it means that if one takes x=0.1 (v=0.1 c) or x=0.999 (v=0.999c), there will be a finite sum for each respective series.
Yes, bit that is irrelevant with regards to the limit as ##x\to 1##. What is relevant for that limit is that this finite value grows without bound the closer you get to x=1.
 
  • #18
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Once we said that the series converges to a finite sum for all abs x < 1, than to me it means that no divergence could take place as x approaches 1.
You expect the sum of the series to be some finite value as long as ##|x|<1##, yes?

So, as ##x \to 1## that value increases. The closer ##x## gets to ##1## the larger the value.

So, indeed, there are no problems finding a finite value as long as ##|x|<1##.

But there is a problem finding a finite value when ##x=1##; the nature of the problem being that there is no finite value.
 
  • #19
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So we have an issue about two different convergence questions: one being about the infinite series as the number of terms n approaches infinity, and one talking about what happens as the value x approaches 1.


Correct?

In any event, the latter diverging seems pretty intuitive as you are approaching division by zero when x approaches 1, and the smaller the denominator the larger the fraction in general, and there seems to be nothing in this case that would limit the size.


Does this sound fairly close to right? I’ve been following along here and want to make sure I’ve understood.
 
  • #21
Ibix
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Close to x=1, four terms is nowhere near enough for precision, so your result is nowhere near correct. No time at the moment to do a proper analysis so, back of the envelope, you are going to need at least a hundred terms to get anywhere near precise results for x=0.99. (Edit: commuter trains are quieter than libraries. The rule seems to be that you need approximately ##1.7\times 10^k## terms to get within 1% of ##\left(1/\sqrt{1-x^2}-1\right)## when ##x=0.999\ldots## with ##k## 9s - so x=0.99, you need 170 terms, x=0.999 you need 1700, etc.)

Edit 2: Notice that your own graph is showing you this. Only the black line is beginning to flatten out - so you don't have a good approximation for any of them. I expect you need around 17 terms to get a decent "final" value for the black line. But you need around 1700 to be able to see your upper two lines separating.

The series expansion of ##\gamma## is usually used when x is very close to zero, to show that Newton was approximately correct for v<<c. For that purpose, four terms is more than enough. You are trying to do something else, and I'm not sure what your goal is - to show that Taylor series expansion is wrong, or something?
 
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  • #22
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Say
[tex]\gamma=\frac{1}{\sqrt{1-x^2}}[/tex]
[tex]\gamma^2=\frac{1}{1-x^2}=1+x^2+x^4+x^6+x^8+x^{10}+...[/tex]
[tex]x\rightarrow 1-0,\ \ \gamma^2\rightarrow 1+1+1+1+1+1+...= +\infty, \ \ \gamma\rightarrow +\infty[/tex]
 
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  • #23
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Since I have been working on calculating higher derivatives for another purpose, here are the first six derivatives of ##\frac {1} {\sqrt{1 - v^2}}## evaluated between ##v = 0.1## and ##v = 0.9##.
Key: black = value, red = f', green = f'', blue = f''', yellow = f'''', cyan = f''''', magenta = f''''''.
Figure_1.png
 

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  • #24
Ibix
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Would you be able to demonstrate an expression for the sum of the series?
I sincerely hope it sums to ##1/\sqrt{1-x^2}##, given that that is what we are approximating. It's a standard result - the Taylor series expansion of a binomial can be found in many places. For example, the expansion can be found in the "List of Maclaurin series of some common functions" section of Wikipedia's page in Taylor series. Use ##\alpha=-1/2## (which is actually worked out for you) to get the expansion of ##1/\sqrt{1+x}## and replace ##x## in the result with ##-x^2##.
 
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Hello Ibix,

Thank you for the comments. I am certainly aware that using more terms of the series will provide a more precise outcome. I definitely know that the expansion leads to the Newton formula ar v<<c.
It is not clear to me how closely the expansion series we use, see below, approximates the original, not expanded, KE formula. Would you be able to demonstrate the expression for the sum of the series?

upload_2018-11-2_17-35-44.png
 

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