Problem understanding relativistic momentum

[anantchowdhary]

Many books teaching relativity at the introductory level,go on and say that relativistic momentum can't be described using p=mv

and p=m Dx/Dt(D is Delta)

also

it says the measure Dx is the same as measured by the both people.One who is measuring his own momentum and another who is measuring the moving person's(with respect to him) momentum.

I get confused.how can this be?

 

[Hootenanny]

Many books teaching relativity at the introductory level,go on and say that relativistic momentum can't be described using p=mv

and p=m Dx/Dt(D is Delta)

also

it says the measure Dx is the same as measured by the both people.One who is measuring his own momentum and another who is measuring the moving person's(with respect to him) momentum.

I get confused.how can this be?

Perhaps it is meant to say something like the following;

"The above equations assume that the measure[ment] Dx is the same as measured by the both people.One who is measuring his own momentum and another who is measuring the moving person's(with respect to him) momentum."

 

[anantchowdhary]

Why shd they both be the same?As the person will refer to another thing to measure Dx.Like the length of road he covers.But the road will seem contracted to him isn't it?

 

[neutrino]

May be quoting the text would help clear things up.

 

[Hootenanny]

Why shd they both be the same?As the person will refer to another thing to measure Dx.Like the length of road he covers.But the road will seem contracted to him isn't it?

They are not the same in special relativity, the quantity dx will indeed be subject to length contraction as described by the lorentz transformations. However, the two equations you have quoted are the equations for classical momentum which are based on the Galilean transformations and hence dx is not subject to length contraction.

 

[anantchowdhary]

Heres the text quoted:
Consider a particle moving with a constant speed v in the positive x direction.Classically it has momentum=mv=mDx/Dt

in which Dx is the distance it travels in time Dt.To find a relativistic expression for momentum,we start with the new definition

p=mDx/Dt(o)

Here,as before Dx is the distance traveled by a moving particle as viewed by the observer watching that particle.However t(o) is the time required to travel that distance,measured by not the observer watching the moving particle but by an observer moving with the particle.

Using time dilation t/gamma=t(o)

therefore p=mv[gamma]


I didnt understand why distance Dx is same for both observers!

 

[Doc Al]

I didnt understand why distance Dx is same for both observers!

Where in that quoted passage does it say that Dx is the same for both observers? It says: "Dx is the distance traveled by a moving particle as viewed by the observer watching that particle". Nowhere does it say that different observers measure the same distance.

 

[anantchowdhary]

How is the textbook then comparing the two momentum(S) measured by the person moving with the particle and the person who is not at rest with respect to the particle?

 

[Hootenanny]

How is the textbook then comparing the two momentum(S) measured by the person moving with the particle and the person who is not at rest with respect to the particle?

Time dilation, note the term t₀, the proper time interval.

 

[neutrino]

The argument as per the book:

Rest Frame:

$$p_{0} = m\frac{\Delta x}{\Delta t_0}$$

Moving frame:

$$\Delta t = \gamma\Delta t_0$$

$$\therfore p_{0} = \gamma m\frac{\Delta x}{\Delta t}$$

But shouldn't it be...

Rest Frame:

$$p_{0} = m\frac{\Delta x_0}{\Delta t_0}$$

Moving frame:

$$\Delta t = \gamma\Delta t_0$$

$$\Delta x = \gamma^{-1}\Delta x_0$$, and therfore

$$p = m\frac{\Delta x}{\gamma^2\Delta t}$$

?

 

[Hootenanny]

But shouldn't it be...
Rest Frame:
$$p_{0} = m\frac{\Delta x_0}{{\color{red}\Delta t_0}}$$

Are you sure about that? Read again, t₀ is not in the rest frame of the observer, but in that of the particle.

 

[neutrino]

Oh, okay. So from the particle's frame only length contraction is observed. So that would lead to only one gamma factor.

 

[anantchowdhary]

im confused
Can som1 please explain well.I am asorry I've tried a lot but cudnt understand.Why would $$\Delta x$$ be the same for both?

 

[Hootenanny]

Oh, okay. So from the particle's frame only length contraction is observed. So that would lead to only one gamma factor.

Correct. Indeed, one can do equivalent procedures by considering length contraction or time dilation. The particle experiences length contraction, whereas the observer experiences time dilation; both will lead to the same results, provided the principles are applied correctly.

im confused
Can som1 please explain well.I am asorry I've tried a lot but cudnt understand.Why would $$\Delta x$$ be the same for both?

No one has ever said they are! $\Delta x$ is the distance measured by the observer not by the particle.

 

[anantchowdhary]

Explanation please.I still can't get it :'(

 

[neutrino]

Thanks, Hoot.

 

[Hootenanny]

Explanation please.I still can't get it :'(

What exactly don't you understand?

Thanks, Hoot.

No problem

 

[anantchowdhary]

Why length contraction isn't observed!Also which part of neutrino's equations were correct?Thnx

 

[Hootenanny]

Why length contraction isn't observed!Also which part of neutrino's equations were correct?Thnx

Length contraction is observed, just not by an observer.

Correct. Indeed, one can do equivalent procedures by considering length contraction or time dilation. The particle experiences length contraction, whereas the observer experiences time dilation; both will lead to the same results, provided the principles are applied correctly.

Also which part of neutrino's equations were correct?Thnx

The first part under the heading 'According to the Book' is correct.

 

[anantchowdhary]

O boy i just can't seem to get it

According to me:

We are comparing the momentum an observer measures of a moving body and which the moving body measures of itself.Now suppose the moving body is on a road and has two stones which have a distance of Dx(with respect to the observing frame,i mean the rest observer).

Now when the observer measures the body's momentum:
$$p = m\frac{\Delta x}{\Delta t}$$ ...(1)

also when the moving body measures his own momentum
$$p_{0} = m\frac{\Delta x_0}{\Delta t_0}$$
but as
$$\Delta x_0 = \gamma^{-1}\Delta x$$

Now I am getting:
$$p_{0} = m\frac{\Delta x}{\gamma^2\Delta t}$$

So that's where I am going wrong!:'((((((((((((((

 

[Hootenanny]

Now when the observer measures the body's momentum:
$$p = m\frac{{\color{red}\Delta x}}{\Delta t}$$ ...(1)

This length is a proper length, since it is measured in the rest frame of your two stones.

also when the moving body measures his own momentum
$$p_{0} = m\frac{{\color{red}\Delta x_0}}{\Delta t_0}$$

The length measure by the moving body is not a proper length. In your example above, the particle is moving relative to the distance between the two stones, therefore the distance between the two stones appears contracted!

 

[anantchowdhary]

err i think i said that$$\Delta x_0 = \gamma^{-1}\Delta x$$

and $$p_{0}$$ is the momentum as measured by the moving body

 

[Hootenanny]

err i think i said that$$\Delta x_0 = \gamma^{-1}\Delta x$$

and $$p_{0}$$ is the momentum as measured by the moving body

You did indeed say that. However, the two statements I highlighted are still incorrect for the reasons I stated above.

 

[anantchowdhary]

err... how?

 

[Hootenanny]

err... how?

Are you even reading my posts?

 

[anantchowdhary]

I am reading ur posts.But X(o) is the distance measured by the body in motion.Why isn't that correct because its lesser than x

 

[neutrino]

If you're moving with the particle, you'll measure a certain distance between the stones, say L. But this L is equal to \gamma^-1.L_0, where L_0 is the distance between the stones in the stones' rest frame.

For someone in the stones' rest frame, the distance will not be contracted, but your clocks rate is slower relative to his. T = T_0\gamma, where T_0 is the time you took the cross the L_0 as measured by this observer.

 

[Hootenanny]

I am reading ur posts.But X(o) is the distance measured by the body in motion.

No it isn't! x₀ is the proper length. Look up the definition of proper length. The proper length is the length measure between two events (or stones) in their rest frame and the particle definatly isn't in the rest frame of your two stones because its moving past them!

An alternative definition;
The proper length is the difference between the spatial coordinates of two events (or stones) in a frame of where the two events appear simultaneous.

 

[anantchowdhary]

im taking x as the proper length.So x0 is lesser than x

 

[Hootenanny]

im taking x as the proper length.So x0 is lesser than x

Perhaps its time someone else stepped into help you...