Relativistic Density - Check My Work & Learn Methods

In summary: My brain hurts.Can you compare your formula for length contraction to the one in the book? When length "contracts", does it get larger or smaller?I am also concerned about your units. Your contracted length is expressed in kilograms?Or is density supposed to be unchanged??My brain hurts.
  • #1
LilRubyKinz
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Homework Statement
Consider a cube to have a density of 2.0kg/m^3. What is its relativistic density at 0.95c?
Relevant Equations
Lm = Ls/(1-v^2/c^2)^ - 1/2
mm = ms/(1-v^2/c^2)^ - 1/2
Density = mass/volume
We never learned how to use these formulas, so I'm pretty much grasping for straws. I got 2.04 kg/m^3. Can anyone check my work/teach me proper methods if I'm wrong? Thanks.
 
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  • #2
Wait that's definitely wrong... Now I'm getting 42.3. Is that right?
 
  • #3
LilRubyKinz said:
Wait that's definitely wrong... Now I'm getting 42.3. Is that right?

Can you explain what you're doing?
 
  • #4
Yes. I'll attach a picture.
 

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  • #5
I wrote the last division statement wrong (switch numerator and denominator) but yes I get 42.3 or 45 if I don't round.
 
  • #6
LilRubyKinz said:
Homework Statement: Consider a cube to have a density of 2.0kg/m^3. What is its relativistic density at 0.95c?
Homework Equations: Lm = Ls/(1-v^2/c^2)^ - 1/2
mm = ms/(1-v^2/c^2)^ - 1/2
Density = mass/volume
Typesetting those equations for you. See this link to see how it was done.

$$L_m = \frac{L_s}{\sqrt{1-v^2/c^2}}$$
$$m_m = \frac{m_s}{\sqrt{1-v^2/c^2}}$$

One assumes that ##L_m## is the length of the moving cube and ##L_s## is its length as measured from a frame where it is stationary. Similarly, ##m_m## would be the relativistic mass of the moving cube and ##m_s## is its mass as measured from a frame where it is stationary.

Did you forget to square something?
 
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  • #7
If I can understand what you're doing, you are length contracting all three dimensions of of the cube.
 
  • #8
PeroK said:
If I can understand what you're doing, you are length contracting all three dimensions of of the cube.
Yes I think that's what I'm doing. Am I not supposed to? Is this the wrong formula?
 
  • #9
Length contraction is only in the direction of relative motion.
 
  • #10
So what formula should I be using to get the new volume?
 
  • #11
LilRubyKinz said:
So what formula should I be using to get the new volume?
Same one they taught you in grade school. Volume = length * width * height.
 
  • #12
Okay, so I don't need to change anything about length? Just mass changes?
 
  • #13
LilRubyKinz said:
Okay, so I don't need to change anything about length? Just mass changes?
What makes you think that length contraction does not apply?
 
  • #14
jbriggs444 said:
Length contraction is only in the direction of relative motion.

I confuse easily, sorry. I thought this meant not to use it.

Can you please walk me through the correct process of calculating the new mass and lengths?
 
  • #15
LilRubyKinz said:
I confuse easily, sorry. I thought this meant not to use it.

Can you please walk me through the correct process of calculating the new mass and lengths?
You already quoted the correct formula for length contraction. By what factor has the length of the cube contracted?

This factor is also known as gamma (##\gamma##).
 
  • #16
Okay here's the problem. The square root of 1 - (0.95c)^2 / c^2 should be 0.312. I calculated this wrong.

I'm going to try again with this new value; I'll let you know what I get
 
  • #17
LilRubyKinz said:
Okay here's the problem. The square root of 1 - (0.95c)^2 / c^2 should be 0.312. I calculated this wrong.

I'm going to try again with this new value; I'll let you know what I get
Put your calculator away and start thinking instead!
 
  • #18
Is only one side length affected by length contraction or all three?
 
  • #19
LilRubyKinz said:
Is only one side length affected by length contraction or all three?

Who says the cube is moving parallel to one of its edges? But, let's assume it is, what do you think?
 
  • #20
Yes?
 
  • #21
LilRubyKinz said:
Yes?

There is no length contraction in directions perpendicular to the motion.
 
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  • #22
PeroK said:
There is no length contraction in directions perpendicular to the motion.

So does that mean I'm right? Sorry I'm not great at this!
 
  • #23
LilRubyKinz said:
So does that mean I'm right? Sorry I'm not great at this!
Right about what?
 
  • #24
LilRubyKinz said:
So does that mean I'm right? Sorry I'm not great at this!

Yes, only one side is contracted.
 
  • #25
PeroK said:
Yes, only one side is contracted.

Okay thank you!

Now one more thing I don't understand.

mm = 2 / 0.312 = 6.4 m
Lm = 1 / 0.312 = 3.2 kg

6.4 / 3.2(1)(1) = 2

Why am I still getting 2 as the answer? Does the density not change?
 
  • #26
LilRubyKinz said:
Okay thank you!

Now one more thing I don't understand.

mm = 2 / 0.312 = 6.4 m
Lm = 1 / 0.312 = 3.2 kg

6.4 / 3.2(1)(1) = 2

Why am I still getting 2 as the answer? Does the density not change?
Can you compare your formula for length contraction to the one in the book? When length "contracts", does it get larger or smaller?

I am also concerned about your units. Your contracted length is expressed in kilograms?
 
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  • #27
So I am supposed to multiply there, not divide? Making the new length 0.312?

Making the answer 20.5?

Or is density supposed to be unchanged??

My brain hurts.
 
  • #28
jbriggs444 said:
Can you compare your formula for length contraction to the one in the book? When length "contracts", does it get larger or smaller?

I am also concerned about your units. Your contracted length is expressed in kilograms?

Sorry, swap kg and m there. Copied it wrong.

But the book shows Lm and Ls reversed from what I have here...

So 20.5 should be the answer? Or 2?
 
  • #29
Go back to basics.

Moving objects appear to be shortened. Short objects have lower volume. Lower volume means higher density.

Moving objects gain relativistic mass. [Note that relativistic mass is a deprecated concept that would better be left untaught]. More mass means higher density.

Both effects call for an increased density. So an answer of 2.0 kg/m3 could not be correct.

20.5 kg/m3 matches what I get. Though you might want to consider rounding to an appropriate number of significant figures.
 
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  • #30
jbriggs444 said:
Both effects call for an increased density. So an answer of 2.0 kg/m3 could not be correct.

Okay, getting a little bit of mixed messages here... But I think 20.5 makes the most sense to me. Thank you very much! I will fix significant digits now.
 
  • #31
I think @PeroK and myself are on the same page here. If you perceive a conflict, either of us can probably clarify it away.
 
  • #32
Okay thank you!
 
  • #33
jbriggs444 said:
I think @PeroK and myself are on the same page here. If you perceive a conflict, either of us can probably clarify it away.

Yes, I agree.

I was distracted trying to find a way to show that it's true in general, even when motion is not in the same direction as one side of the cube. Which I've just spotted!
 
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  • #34
PeroK said:
I was distracted trying to find a clever way to show that it's true in general, even when motion is not in the same direction as one side of the cube. Which I've just spotted!
I was taking it as an obvious geometric property -- scale down one dimension by a factor of ##\gamma## and the volume clearly goes down proportionately.

I gave a moment's though to Terrell rotation, but it does not apply.
 
  • #35
jbriggs444 said:
I was taking it as an obvious geometric property -- scale down one dimension by a factor of ##\gamma## and the volume clearly goes down proportionately.

Yes, I was only thinking about cubes, parallelopipeds and the triple scalar product! Then, I realized ...

o:)
 

1. What is relativistic density?

Relativistic density is a concept in physics that describes the density of an object or a system in terms of its mass and volume, taking into account the effects of special relativity. It is a measure of how much mass is packed into a given volume, and it changes as the object's velocity approaches the speed of light.

2. How is relativistic density calculated?

Relativistic density can be calculated using the formula ρ = m/γV, where ρ is the relativistic density, m is the rest mass of the object, V is its volume, and γ is the Lorentz factor, which takes into account the effects of special relativity.

3. Why is it important to consider relativistic density?

Relativistic density is important because it helps us understand the behavior of objects and systems at high velocities. As an object's velocity increases, its relativistic density also increases, which can have significant effects on its mass, energy, and other properties. This is especially important in fields such as astrophysics and particle physics.

4. How does relativistic density differ from classical density?

Relativistic density differs from classical density in that it takes into account the effects of special relativity, while classical density does not. This means that relativistic density can change as an object's velocity changes, while classical density remains constant regardless of velocity.

5. What are some methods for calculating relativistic density?

There are several methods for calculating relativistic density, including using the Lorentz factor, the relativistic mass formula, and the relativistic energy formula. Other methods include using the Schwarzschild radius and the density formula for a Schwarzschild black hole. These methods may vary depending on the specific situation and the level of accuracy required.

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