SR Limit Problem: mcc/sqrt(1-uu/cc)

  • Thread starter Thread starter bernhard.rothenstein
  • Start date Start date
  • Tags Tags
    Limit Sr
bernhard.rothenstein
Messages
991
Reaction score
1
Consider the limit of the function

mcc/sqrt(1-uu/cc)
for u=c. (m rest mass, u speed of the particle with rest mass m relative to a given inertial reference frame.

Please tell me what is its physical meaning. Do you see there a connection with the energy of a photon?
Thanks in advance for your answer.
 
Physics news on Phys.org
Here are some hints:
\frac{m c^2}{\sqrt{1 - \frac{u^2}{c^2}}} = \gamma m c^2 = E_0
does that equation look familiar? What happens to \gamma \text{ as } u \to c?
What is the energy of a photon?

Now please show us some attempt (please try thinking for yourself before asking us to do it for you).
 
Last edited:
limit

CompuChip said:
Here are some hints:
\frac{m c^2}{\sqrt{1 - \frac{u^2}{c^2}}} = \gamma m c^2 = E_0
does that equation look familiar? What happens to \gamma \text{ as } u \to c?
What is the energy of a photon?

Now please show us some attempt (please try thinking for yourself before asking us to do it for you).
Is your hint correct?
 
When taking limits, there are usually constraints under which the limit is taken... often they are implicit and obvious. For this question, can you make those constraints explicit?

For instance, What is m? (i now see it above) and (more importantly) is it held fixed while taking this limit?
 
Last edited:
bernhard.rothenstein said:
Is your hint correct?

Isn't it?
If I made a mistake, please tell me -- it happens sometimes (actually, it happens all the time).
 
relativistic limit

CompuChip said:
Isn't it?
If I made a mistake, please tell me -- it happens sometimes (actually, it happens all the time).
Cut the index zero at the last E.
 
relativistic limit

robphy said:
When taking limits, there are usually constraints under which the limit is taken... often they are implicit and obvious. For this question, can you make those constraints explicit?

For instance, What is m? (i now see it above) and (more importantly) is it held fixed while taking this limit?

Thanks.
In short my problem is:
Start with the expression for the kinetic energy of a tardyon
W=mcc/sqrt(1-uu/cc)-mcc
with m for the rest mass.
Apply it in the case of a photon (u=c, m=0).
W(photon)=0/0.
Is it correct to extend that result to
W(photon)=hniu
taking into account that the energy of a photon is kinetic in its essence?
Please consider that all I say are questions and not statements.

I
 
Consider the limit of the function

mcc/sqrt(1-uu/cc)
for u=c. (m rest mass, u speed of the particle with rest mass m relative to a given inertial reference frame.

Please tell me what is its physical meaning. Do you see there a connection with the energy of a photon?
Thanks in advance for your answer.
Consider how one usually derives that expression, i.e. from the Work-Energy Theorem. For those who are rusty on this I worked it out for relativity on one of my web pages. I'm sure the math could have been streamlined better and will do so when I find a shorter/cleaner derivation. And advice on how to do that is welcome!

The derivation starts with a particle at rest, such as a charged particle which is initially at rest in the inertial frame S. A force is then applied to the particle, e.g. an electric field is established, and the particle begins to accerate, the final velocity then being v (or u as you refer to it as). This derivation assumes that the particle can be at rest and that it can accelerate. Neither of which apply to a photon. I recall a limit process in a book by Thorne et al. See - http://www.pma.caltech.edu/Courses/ph136/yr2002/chap01/0201.2.pdf

See page 16 right after Eq. (1.31)

CompuChip said:
Now please show us some attempt (please try thinking for yourself before asking us to do it for you).
The question asked is not a homework problem and as such we don't need to see any attempt by Bernhard to respond to this question. I believe Bernhard is seeking the impressions others have on this concept. Bernhard is a very sharp man. In fact he was a physicist professor before he retired.

Pete
 
limit in special relativity

Thanks Pete
My problem has a more general character.
If I start with the formula which accounts for the longitudinal Doppler effect involving a machine gun which emits bullets at a constant period the bullets hitting a moving target at a period T' then
T'=(1+V/u)/sqrt(1-VV/cc) (1)
where u stands for the speed of the bullet and V for the speed of the target.
If I replace the bullets with light signals (u=c) (1) becomes
T'=(1+V/c)/sqrt(1-VV/cc)
which accounts for the Doppler effect in the optical domain. Do you think that is by chance or there is some physics behind.
In general how does special relativity sccount for the transition from the bullet to the photon?
I learned long time ago that "Natura non facit saltus" ensuring smooth transitions.
Regards
 
  • #10
pmb_phy said:
The question asked is not a homework problem and as such we don't need to see any attempt by Bernhard to respond to this question.
OK, it could have been a homework question, and in general I like to see what the OP already did/tried. For one, it gives me an impression of the level of the poster, so that I don't go around explaining too basic things to someone who long knows them, or start throwing stuff at him/her which is far above the level required. So basically, precisely to avoid this situation :smile:

pmb_phy said:
I believe Bernhard is seeking the impressions others have on this concept.
... in which case my response was probably not very useful. Apparently from time to time I take things for granted too quickly, and at the time the question seemed rather easy (I thought the implied meaning was, that when taking the limit for a massless object going to the speed of light, it's rest energy becomes infinite, referred to as becoming "infinitely heavy", etc. -- which by the way seems to me typically like a homework question, another reason for my hasty assumption). Reading Bernhard's last post, I realize the question is not nearly as trivial as I at first thought, and I probably misunderstood it the first time as well.

pmb_phy said:
Bernhard is a very sharp man. In fact he was a physicist professor before he retired.
One of the dangers of communicating through written text... you never know who's on the other side of the line, do you. In this case, I think I'd better shut up because if you're right Pete, there's probably more I can learn from Bernhard than the other way around :smile:
Bernhard, I hope I haven't offended you (and if by accident I did, that you will accept my apologies).
 
  • #11
kinetic energy in special relativity

pmb_phy said:
Consider how one usually derives that expression, i.e. from the Work-Energy Theorem. For those who are rusty on this I worked it out for relativity on one of my web pages. I'm sure the math could have been streamlined better and will do so when I find a shorter/cleaner derivation. And advice on how to do that is welcome!

The derivation starts with a particle at rest, such as a charged particle which is initially at rest in the inertial frame S. A force is then applied to the particle, e.g. an electric field is established, and the particle begins to accerate, the final velocity then being v (or u as you refer to it as). This derivation assumes that the particle can be at rest and that it can accelerate. Neither of which apply to a photon. I recall a limit process in a book by Thorne et al. See - http://www.pma.caltech.edu/Courses/ph136/yr2002/chap01/0201.2.pdf

See page 16 right after Eq. (1.31)

Thanks
My problem is if relativistic kinetic energy (W) derivation should compulsory involve the kinetic energy-work theorem. I think that in order to save time
in teaching we could start with the expression for the relativistic energy E
(which can be derived without using conservation laws) and to state that the energy E is the sum between kinetic energy and proper energy i.e
W=E-E(0)= E(0)[g(u)-1].
Would you teach the stuff that way?
 
Back
Top