Can Special Relativity's Equations Be Modified to Avoid Infinities and Zeroes?

[kurious]

Is it impossible to accelerate a rest mass to the speed
of light or does special relativity have an infinity that
should be removed.For example, what if

rest mass / (1 - v^2 / c^2)^1/2

should really be rest mass / (1 - v^2 / c^2 + small constant)^1/2

The small constant would also stop lengths contracting to zero.

 

[pmb_phy]

Is it impossible to accelerate a rest mass to the speed
of light or does special relativity have an infinity that
should be removed.For example, what if

rest mass / (1 - v^2 / c^2)^1/2

should really be rest mass / (1 - v^2 / c^2 + small constant)^1/2

The small constant would also stop lengths contracting to zero.

I've never been sure how to respond to a question like this where someone asks if nature should be a different way than it appears to be.

Its all pretty straight forward. If you want to know what "really" is then you start with the postulates and then go from there. The postulates of SR, coupled with the other laws of nature (i.e. energy conservation, Maxwell's equations etc.). With that as a start Einstein showed that the amount of work, W, done to accelerate a particle of proper mass m₀ to a speed v is given by the relation

$$W = m_{0}c^{2} \left ( \frac{1}{\sqrt{1-v^2/c^2}} - 1 \right )$$

The momentum, p, of a particle of proper mass m₀ moving at speed v is given by

$$\bold p = \frac{m_{0}\bold v}{\sqrt{1-v^2/c^2}}$$

Since both W and p are both infinite when v = c it follows that there would be a violation of the principles of the conservation of energy and conservation of momentum for the particle to move that fast since if the particle started out at rest that energy and momentum had to come from somewhere.

Pete

 

[kurious]

Surely all particles - even photons- must be accelerated up to the speed they have -
afterall some photons are made of two rest masses - positrons and electrons fpr example.

 

[pmb_phy]

Surely all particles - even photons- must be accelerated up to the speed they have -
afterall some photons are made of two rest masses - positrons and electrons fpr example.

Photons can't be accelerted in an inertial frame. But what does this have to do with your question? Thanks

Pete

 

[kurious]

The point I'm making is that it might take a large but finite energy to accelerate a particle with rest mass to the speed of light.Infinity is not accepted in quantum field theory why is it accepted in special relativity - infinity is not a number.

 

[pmb_phy]

The point I'm making is that it might take a large but finite energy to accelerate a particle with rest mass to the speed of light.Infinity is not accepted in quantum field theory why is it accepted in special relativity - infinity is not a number.

I understand your point. You seem to have the impression that infinity is a number. It is not a number. You also seem to think that infinite energy/momentum is something allowable in SR. It is not allowable.When a physicists poses a question then he uses the laws of physics, expressed in the form of equations, to answer them. When the result of such an answer is "infinite mass/momentum etc" then the way to interpret that is that one or more of the starting assumptions are wrong. The starting assumption that is flawed is assuming that a particle can have the speed v. So when the answer is infinite then it means "Can't be done".

The answer to your question "should really be rest mass / (1 - v^2 / c^2 + small constant)^1/2" the answer is no.

Pete

 

[kurious]

The small constant would also stop lengths contracting to zero.

Wouldn't that prevent the formation of a singularity in black holes
and at the beginning of the universe?
And ignoring the fact that you don't agree with having the small constant in the first place,if it did exist,what effects would it have on the theory of general relativity?

 

[woodysooner]

good pt.

I am no pro at any of this, but I have always just taken speed of light and never thought too much about it but,

Surely all particles - even photons- must be accelerated up to the speed they have -

if you have a flashlight and you turn it on, i can't see how instintaneously the photons could be moving the speed of light, i know i have been taight they always move at the speed of light which i can accept but, logically thinking for any thing to get from point a to point b in a certain amount of time from rest they must start going and if they want to speed up to get to a certain limit such as c you would get a vel then an acceleation by def of the next derivative.

I know DW can explain this well he's brilliant, how can you go from rest in flashlight to C without first a increasing vel causing acc. only way would be to travel linearly to the speed of light which would take hellalong time, well even that would have acc but it would be constant.

DW i await your elegant solution.

 

[DW]

if you have a flashlight and you turn it on, i can't see how instintaneously the photons could be moving the speed of light, i know i have been taight they always move at the speed of light which i can accept but, logically thinking for any thing to get from point a to point b in a certain amount of time from rest they must start going and if they want to speed up to get to a certain limit such as c you would get a vel then an acceleation by def of the next derivative.

I know DW can explain this well he's brilliant, how can you go from rest in flashlight to C without first a increasing vel causing acc. only way would be to travel linearly to the speed of light which would take hellalong time, well even that would have acc but it would be constant.

DW i await your elegant solution.

The solution is simple. You are making the assumption that before turning on the light there are photons sitting at rest with respect to the flashlight which then speed up in the process of emission. That is a false assumption. The photons in the beam didn't exist prior to emission. They are created with an initial velocity of c, so there was no acceleration process at all. Nothing massless can travel at any other speed than c.

 

[woodysooner]

interesting, I like it, but how was that proved, or how is it now proved, is itjust because all massless particles have V at C at all pts in time, and if so how is that prooved.

 

[Atheist]

interesting, I like it, but how was that proved, or how is it now proved, is itjust because all massless particles have V at C at all pts in time, and if so how is that prooved.

EDITED VERSION (thx@DW 4 help):
----------------------------------

Dispersion relation for free particles:
$$E(p) = \sqrt{ p^2c^2 + m^2c^4}$$ (1)


Wave function of a free particle:
$$\Psi = C \cdot \exp \left[ \frac{-i}{\hbar} \left( Et - px \right) \right]$$ (2)

Renaming
$$\frac{E}{\hbar} =: \omega, \, \frac{p}{\hbar} =: k$$ (*)
leads to:
$$\Psi = C \cdot \exp \left[ -i \left( \omega t - kx \right) \right]$$ (2*)
Which is the classical wave-equation in the variables usually used.

Velocity of a wave is:
$$v = \frac{d\omega}{dk} \stackrel{(*)}{=} \frac{dE}{dp}$$ (3)

Plugging in the dispersion relation (1) in (3) results in:
$$v = \frac{dE}{dp} = \frac{d}{dp}\sqrt{ p^2c^2 + m^2c^4} = \frac{pc}{E(p)}c$$ (4)

From (4) using (1) one can see that v=c <=> m=0 and v<c <=> m>0 (possible problems at m=p=0 not considered)

That would be my guess for a proof. There are, of course, several points where further investigation would be nessecary to make it a complete proof (what happenes when I take solutions of the Dirac-Equation instead of the Klein-Gordon-eq, for example) but I like it as it is.

 

[woodysooner]

$$E(p) = \sqrt{ p^2c^2 + m^2c^4}$$ (1)
So for photons with m=0:
$$E = cp$$ (1*)

Wave function of a free particle:
$$\Psi = C \cdot \exp \left[ \frac{i}{\hbar} \left( Et - px \right) \right] = C \cdot \exp \left[ i \left( \omega t - kx \right) \right]$$ (2)

Velocity of a wave is:
$$v = \frac{\omega}{k} = \frac{E}{p}$$ (3)
note that 2nd "=" comes from (2).

(1*) in (3) => v = c.

Ok wow this is over me but I'm going to try cause I asked for it right. so first thanks athiest a lot for this, your getting me started.

Ok I follwed the beginning till you got to wave function and I didin't see how that fit in. Also the velocity of wave I didn't see how that came about and how it was equal to E/P, i believe you and that it is, but still a lil confused,
and last I know no one will want to do this but can someone tell me what some of those variables are, like in EM right now every variable is used for different things, seems like every chapter has a v but for different things, but if someone can tell me what i , k, x, exp are.

thanx

 

[Atheist]

i : Imaginary unit (-> imaginary numbers, complex numbers)
k: k and omega are p and E divided by h-bar respectively => omega/k = E/p, because h-bar cancels out.
x : Position in space.
exp: Exponential function.

> How the wave-function fit in:
All particles are described by wave-functions.

> Where the formula v = omega/k comes from:
I remebered having learned something like that during my first semesters of studying, so I looked up the formula in book (dunno the english term for those types of books: "equation collection" would be the direct translation from german) and found it there. No explanation given there, though.

EDIT: Flaw i complained about is probably found.

 

[woodysooner]

has is there an imaginary in the wave function, does not the wave function show the probability of a particle to be someone at a certain time, how could the root of a negative number survive in that.

Still not understnad the correlation of the h bar and wave function to the e/p sorry, i try to followyou and can't see it.

 

[DW]

Something seems wrong with what I posted, though. Because when I plug (1) in (3) it turns out that v>c for m>0 because E²>p²c² ... *grmbl

It is correct, it is just that for massive particles v is not $$\frac{\omega }{k}$$. It is $$v = \frac{d\omega }{dk}$$ or in terms of a monochromatic wave state one can define the quantum frequency such that $$v = \frac{c^{2}k}{\omega }$$.

 

[woodysooner]

if v = w/k and for massive objects we take dw/dk would that not be dv with respect to somehting not sure to me that's an acceleration am i wrong?? probably as always

what does just plain K stand for

and what is a monochromatic wave.

thanx

 

[DW]

if v = w/k and for massive objects we take dw/dk would that not be dv with respect to somehting not sure to me that's an acceleration am i wrong?? probably as always

v does not equal w/k for massive objects. v is dw/dk.

what does just plain K stand for

k unbold is 2*pi divided by the wavelength.

and what is a monochromatic wave.

A wave comprised of a single wavelength.

 

[woodysooner]

ok so is v = w/k for small and then derivative of that for massive objects.

 

[DW]

ok so is v = w/k for small and then derivative of that for massive objects.

$$v = \frac{\omega }{k}$$ is only the "special case" for light speed waves. In all cases in general it is $$v = \frac{d\omega }{dk }$$. It just happens that at v = c we get $$\frac{d\omega }{dk } = \frac{\omega }{k}$$. Here is one way to go about it. Start by defining a quantum frequency in terms of the wave number by $$\omega ^{2} = c^{2}k^{2} + constant$$ The constant will turn out to be proportional to the square of the mass, but for now let it be arbitrary. From that equation, find $$\frac{d\omega }{dk}$$. You will get:
$$\frac{d\omega }{dk} = \frac{c^{2}k}{\omega}$$
For a light speed particle $$v = \frac{\omega }{k} = c$$ and inserted into the above results in
$$\frac{d\omega }{dk} = \frac{c^{2}k}{\omega} = c$$
For all particles light speed or not, the numerator is proportional to the momentum and the denominator is proportional to the energy resulting in
$$\frac{d\omega }{dk} = \frac{pc^{2}}{E}$$
Again, for light like particles E = pc and that yields v = c, otherwise its results are consistent with
$$p = \gamma mv$$ and $$E = \gamma mc^{2}$$.

 

[Atheist]

Thx 4 help

It is correct, it is just that for massive particles v is not $$\frac{\omega }{k}$$. It is $$v = \frac{d\omega }{dk}$$ or in terms of a monochromatic wave state one can define the quantum frequency such that $$v = \frac{c^{2}k}{\omega }$$.

That sounds nice and seems to give good results. Thx a lot! I´ll edit my first post, then.
In case you know that off your head: Could you give a quick explanation why v = dw/dk ? I just remembered having done smth like that some years ago (also remembered there was one velocity with v=w/k and one with v=dw/dk, but I made a mistake when trying dw/dk -forgot factor 0.5 when taking derivativce of sqrt- and came to senseless a result) and just looked that up, but an explanation would be nice.

 

[DW]

That sounds nice and seems to give good results. Thx a lot! I´ll edit my first post, then.
In case you know that off your head: Could you give a quick explanation why v = dw/dk ? I just remembered having done smth like that some years ago (also remembered there was one velocity with v=w/k and one with v=dw/dk, but I made a mistake when trying dw/dk -forgot factor 0.5 when taking derivativce of sqrt- and came to senseless a result) and just looked that up, but an explanation would be nice.

It is the difference between a phase wave velocity and a group wave velocity. If you have a free wave packet it will be comprised of a continuum of phase waves. A single free particle can not be described as a continuum of waves of equal wavelength so its speed can not be given by the speed of a particular phase wave. Instead its speed is given by the behavior of the group. Consider that its phase waves each have a frequency wavelength relationship given by $$\omega ^2 = c^{2}k^{2} + constant$$. Its group velocity will be given by $$v = \frac{d\omega }{dk}$$. Doing the differentiation, for a particle in a monochromatic beam state this yields
$$v = \frac{c^{2}k}{\omega }$$. For a single particle in any continuum state this then would be representative of an expectancy velocity given an expectancy frequency and an expectancy wavelength. That the group velocity is given by $$\frac{d\omega }{dk}$$ is due to the group velocity being a composition of the phase wave velocities.

 

[woodysooner]

DW please be patient with me, i know you will, you have helped me a lot so far, I have had cal I II and III and i don't see how that is the derivative. can you go a lil slower, is it diffE

did you just take der of w^2 of left side the k^2 on the right side.

then set equal to v

i thot you could only have one variable the other's derivative would be 9 since it's a constant.

i know you are right can you explain.

 

[DW]

DW please be patient with me, i know you will, you have helped me a lot so far, I have had cal I II and III and i don't see how that is the derivative. can you go a lil slower, is it diffE

did you just take der of w^2 of left side the k^2 on the right side.

then set equal to v

i thot you could only have one variable the other's derivative would be 9 since it's a constant.

i know you are right can you explain.

Both $$\omega$$ and k are the variables in the frequency-wavelength relation$$\omega ^2 = c^{2}k^2 + constant$$ Differentiate with respect to k to arrive at
$$2\omega \frac{d\omega}{dk} = 2c^{2}k + 0$$. Simplified:
$$v_{g} = \frac{d\omega }{dk} = \frac{c^{2}k}{\omega }$$
Defining momentum and energy in terms of frequency and wavelength then results in
$$v = \frac{c^{2}p}{E}$$
Granted, the expectancy frequency and wavelength are "constants of the motion" just as energy and momentum are, they are not treated as constants in the differentiation because you are looking at how they are related differentially. It is their differential relationship that gives the velocity even when they are held constant.

 

[woodysooner]

thanx so much dw, can you give me the link to your site, i want to check it out maybe i can learn soemthing from it.

 

[DW]

thanx so much dw, can you give me the link to your site, i want to check it out maybe i can learn soemthing from it.

I've gone into more detail about this here than at my site so what I'll do is give you a link to a couple of pages that goes on from there to demonstrate a lot more detail conserning relativistic dynamics.
http://www.geocities.com/zcphysicsms/chap3.htm
http://www.geocities.com/zcphysicsms/chap5.htm

 

[Gerinski]

Hi, I'm new user, just an afficionado to popular science but not skilled in real physics or math. I posted a new thread a few days ago which got no replies so far, and I see this one has something to do with my question, so I post again my question here. For what I read it feels like DW could help ! (pls not too much math)

As I understand it, we may consider our universe split in 2 quite different parts: the universe of matter and the universe of radiation.
Radiation (in vacuum) is 1-dimensional, distances and time shrinked to zero size (if I'm correct, actually space dimension in the direction of your movement is shrinked to zero and time dimension is extended to infinite, which in fact amounts to much the same as being shrinked to zero as well).
i.e. when we picture light in a light-cone we draw a line at 45º but actually all of its points are a single point (the photon leaving the sun and the Earth 8 minutes later are actually "the same place at the same instant" for the radiation). Or in other words, "the extension of the space-time dimensions is perceived as zero" = "there's no perception of the extension of any dimensions").
On the other hand the universe of matter is (at least) 4-dimensional, we experience "change of the 3-dimensional status through time" .

Can we say that the universe we experience is made up by the interaction of the matter 4-dimesional universe "sweeping through" a miriad of 1-dimensional radiation "point-universes" ?

(I'm aware the term "sweeping through" already conveys the idea of change along the time dimension , so it's surely incorrect to say something like "4-dimensional universe sweeping through...". "3-dimensional universe sweeping through ... as times goes by" is enough). But I hope you get the point

In the end the question boils down to the typical wondering of "how is the universe like from the point of view of radiation?"
If the proceeding of time is stopped and multiple points of space-time (distant in space from our point of view) are occupied "without any time interval between them" by the radiation (again, events which from our point of view would be labeled with different time-stamps), does that mean that every position in space-time is "timelessly linked to every other" ?

Even for the events out of our direct light cones, it seems to me that there can always be other events linking both indirectly. (i.e if we are in spacetime position A, and C is somewhere beyond our (let's say) past light-cone, there may be some event B which reaches within its past and future light-cones both A and B. If so, A and C are anyway indirectly dependant and linked (any event is correlated with every other event, even with those beyond its light-cones).

Is this statement correct?
This is non-locality, and I read that GR really does not allow for it. Only QM (Bell's inequality / Aspect experiment etc) showed that non-locality is real.
Knowing that GR (even Einstein himself) did not believe in non-locality until it was proven experimentally, I wonder what's wrong in my reasoning.

It just seems common sense to me that even if 2 events are too distant in spacetime to have direct light-speed interaction, it's possible to exist some intermediate event which links indirectly both of them (A and C not sharing any area of their light-cones, yet B sharing some light-cones area with both A and C, thus providing the indirect link between A and C).

But I don't think I'm smarter than Einstein & fellows! can someone clarify what's wrong in this reasoning?
Even knowing that Aspect showed later that non-locality is real, I want to understand why Einstein & fellows believed for a long time that it was not. Surely they thought about my argument too but they saw it was incorrect. Why ?

TX !