View Full Version : Speed of Light Question

Christine112

2002-Feb-05, 06:47 PM

This question may be spilling over more into physics, but we were learning about relativity in my astronomy class, and we could not figure out an answer to this question...If you have a car traveling 30 mph and another car traveling the same speed, when they hit, physics tells us that they will hit with the force of something going 60 mph. But if you have something traveling .7 the speed of light that collides with something else going .8 the speed of light, the most they can combine to is the speed of light. Doesn't this demonstrate a loss of momentum or force? Or can it be explained by the fact that they are not mass and just light (since mass could not even reach the speed of light)? I know this question isn't in very technical terms and I've probably used the wrong words to describe what I'm trying to get across...sorry I am only just learning about these concepts...

Wally

2002-Feb-05, 07:03 PM

Since I'm the first one here, I'm going to take a crack. first off, I'm a complete amateur when it comes to GR and SR, but here goes nothing.

As long as nothing exceeds the speed of light, you don't have a problem here. Neither object does, so so far, so good. Now, the energy expended when they hit head on (wow, to see that one!) would take into account the inertial energy of both masses (which accounts for their individual speeds). The energy output would in fact be computable and all mass/energy would be conserved. So again, so far, so good. I think your teacher is just trying to pull a fast one on you. Bottom line, the energy expended when they hit doesn't "pass" the speed of light, so no rules are broken. Ok Grapes o' Wrath. Tear me apart here!

Wally

GrapesOfWrath

2002-Feb-05, 07:33 PM

On 2002-02-05 14:03, Wally wrote:

Bottom line, the energy expended when they hit doesn't "pass" the speed of light, so no rules are broken. Ok Grapes o' Wrath. Tear me apart here!

Except for mixing metaphors there, I guess it's OK. Einstein's neat trick was to assume that nothing could exceed the speed of light, and that the speed of light was measured the same for all observers, and then go from there. So, the way we calculate energy and momentum and all the rest of mechanics is different for special relativity--but it's set up so that it all works out without contradiction. That was real beauty, making two small (large?) assumptions, and developing a new physics from them.

One last thing though, there is a huge amount of discussion around the web about the head-on collision of two cars. If they are going 30 mph, they are approaching each other at 60 mph--but each will experience the collision as if they had each hit a brick wall at 30 mph, more or less. Why? Imagine a mirror on the brick wall. If the brick wall does not give, the collision will look about the same, and the car will crumple about the same.

Bob S.

2002-Feb-05, 07:47 PM

Anybody know what happened to the "Relativity FAQ" page that used to be at http://math.ucr.edu/home/baez/physics/relativity.html ? I thought it dealt with this issue, but now I can't find it. The best link I could find is a copy of Einstein's "Relativity" at http://www.bartleby.com/173/ that has a section that deals with addition of velocities at relativistic speeds at http://www.bartleby.com/173/13.html .

Anyway, it covers Lorentz Transformations that show that the speeds do not simply add. Particle "A" traveling at .7c toward particle "B" traveling at it at .8c does not see it coming at 1.5c but rather much less (something like .96c). Actually this is also true with the cars rushing head-on, but their speeds are so low relative to each other and the ground and any observers, the relativistic effects are negligible.

<font size=-1>[ This Message was edited by: Bob S. on 2002-02-05 14:50 ]</font>

GrapesOfWrath

2002-Feb-05, 07:59 PM

On 2002-02-05 14:47, Bob S. wrote:

Anybody know what happened to the "Relativity FAQ" page that used to be at http://math.ucr.edu/home/baez/physics/relativity.html ?

The Usenet Physics FAQ (http://hjem.get2net.dk/keefe/web/faq.html) says Relativity and Cosmology questions are answered in the Relativity FAQ (http://hjem.get2net.dk/keefe/web/relativity.html). That looks like the same one.

[quote]

But if you have something traveling .7 the speed of light that collides with something else going .8 the speed of light, the most they can combine to is the speed of light.

This statement of initial conditions, while it would be complete in Newtonian mechanics, is incomplete in special relativity mechanics.

According to special relativity (SR), dfferent observers can measure different velocities if the two observers are moving fast with respect to each other. An observer can be a biological entity or a piece of measuring apparatus.

The two velocities given in the problem, 0.7c and 0.8c, are both relative to one observer in an inertial frame. This observer, who is doing experiments with radar and other stuff, measures 0.7c for one object and 0.8c for the other. He also measures momentum and energy of these two objects according to his inertial frame of reference. If he sees them collide, he can state that for his inertial frame they collided at 0.8c+0.7c=1.5c.

He can even calculate the momentum and kinetic energy by use of the Newtonian mechanics.

Doesn't this demonstrate a loss of momentum or force? Or can it be explained by the fact that they are not mass and just light (since mass could not even reach the speed of light)? I know this question isn't in very technical terms and I've probably used the wrong words to describe what I'm trying to get across...sorry I am only just learning about these concepts...

However, he will be dissappointed when he checks out the conservation of momentum (p=mv) and kinetic energy (KE=mv^2/2). The Newtonian laws are inconsistent with the laws of conservation at these high speeds.

One must use special relativistic formulas for momentum and kinetic energy, using concepts such as "longitudinal mass" and "rest mass."

In an case, there is a formula for summing velocities that always results in a sum less than the speed of light. However, the trick here is that there are two observers. One measures 0.8 c of the other observer, the other measures an object 0.7c relative to himself, and the first observer measures and object about 0.9c relative to himself. The trick is that there are actually two observers in the problem.

David Simmons

2002-Feb-06, 03:18 AM

On 2002-02-05 14:33, GrapesOfWrath wrote:

On 2002-02-05 14:03, Wally wrote:

Bottom line, the energy expended when they hit doesn't "pass" the speed of light, so no rules are broken. Ok Grapes o' Wrath. Tear me apart here!

Except for mixing metaphors there, I guess it's OK. Einstein's neat trick was to assume that nothing could exceed the speed of light, and that the speed of light was measured the same for all observers, and then go from there. So, the way we calculate energy and momentum and all the rest of mechanics is different for special relativity--but it's set up so that it all works out without contradiction. That was real beauty, making two small (large?) assumptions, and developing a new physics from them.

One last thing though, there is a huge amount of discussion around the web about the head-on collision of two cars. If they are going 30 mph, they are approaching each other at 60 mph--but each will experience the collision as if they had each hit a brick wall at 30 mph, more or less. Why? Imagine a mirror on the brick wall. If the brick wall does not give, the collision will look about the same, and the car will crumple about the same.

Doesn't the fact that the one traveling at 0.7c has an inertial mass 1.4 times and the one traveling at 0.8c has an inertial mass 1.67 times rest mass rescue the conservation of energy?

Or am I out on a limb here?

<font size=-1>[ This Message was edited by: David Simmons on 2002-02-05 22:21 ]</font>

Bob S.

2002-Feb-06, 03:14 PM

The Usenet Physics FAQ (http://hjem.get2net.dk/keefe/web/faq.html) says Relativity and Cosmology questions are answered in the Relativity FAQ (http://hjem.get2net.dk/keefe/web/relativity.html). That looks like the same one.

Thanks. I figured there had to be more than one copy out there.

Doesn't the fact that the one traveling at 0.7c has an inertial mass 1.4 times and the one traveling at 0.8c has an inertial mass 1.67 times rest mass rescue the conservation of energy?

Or am I out on a limb here?

Doesn't work quite that well. The "longitudinal mass" was defined by Einstein in 1905 to preserve some Newtonian equations, but not all the Newtonian equations.

The phrase, "relativistic mass," usually refers to the "longitudinal mass" given by

m_L=m_0/sqrt(1-{v/c}^2)

where "m_L" is the longitudinal mass, "m_0" is the rest mass, "v" is the velocity relative to the observer,and "c" is the speed of light in a vacuum. A body has a "longitudinal mass." The total energy, "E," of the body is then:

E=(m_L)c^2

The total energy of a system of bodies is conserved.

Let me list some Newtonian formulas that the formula for "longitudinal mass" actually saves even at high velocities.

The body is assumed moving at a constant velocity relative to an observer in an inertial frame. The following Newtonian formulas still work even close to the speed of light for an observer in an inertial frame.

p=m_L v

where "p" is the momentum in the direction of the velocity. The total momentum of a system of bodies is conserved.

Furthermore:

F_L=(m_L) a

where "F_L" is the component of force that is in the direction of thevelocity by the same observer, and a"a" is the acceleration of the body as measured by the observer.

Here are some formulas that the formula for longitudinal mass does not save.

The formula for kinetic energy KE in Newtonian physics:

KE=p^2/(2m)

The formula for kinetic energy in relativity is:

KE=sqrt((cp)^2+(m_0 c^2)^2)-m_0 c^2

The strict replacement of "longitudinal mass" for Newtonian mass doesn't work for kinetic energy. Note that for v<<c, KE(Newtonian) is approximately KE(relativity).

However, it is never exactly the same.

Also, in relativity, mass is anistropic. Mass is a tensor not a scalar. The tranverse mass is different from the longitudinal mass, That is, while

m_L=m_0/(1-{v/c}^2)^0.5

for longitudinal mass "mL,"

(m_T)=(m_0)/(1-{v/c}^2)^1.5

for tranverse mass (m_T).

Then, if the component of force parallel to velocity "v" is "F_L," and the component of acceleration parallel to the velocity is "(a_L)", then:

F_L= m_L (aL)

However, if the component of force perpendicular to velocity "v" is "F_T," and the component of acceleration parallel to the velocity is "(a_T)", then:

F_T= m_T (a_T)

Alot of physicists suggest that one not to use the concept of longitudinal mass, and to hurry right to the momentum-energy relationships. I like longitudinal mass as a concept, although one to be used carefully.

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