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fedorfan

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- Thread starter fedorfan
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fedorfan

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- #2

MeJennifer

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No, it would never reach the speed of light.

- #3

russ_watters

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- #4

AlephZero

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Experiments show that Einstein is more nearly right that Newton was.

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fedorfan

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AlephZero

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That doesn't follow, even for Newtonian mechanics.

Suppose the acceleration is [tex]e^{-t}[/tex] and the initial velocity is 0. The velocity at time t is [tex]1-e^{-t}[/tex]. The accleration is always positive, but the velocity is always less than 1.

- #7

electricsheep

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The faster something moves, the more kinetic energy it has. E=mc^2, so the faster it moves, the heavier it becomes. The heavier it becomes, the greater force is required to accelerate it more. If that is possible, then it becomes faster and hence heavier again. As you can see, it's just a vicious cycle, meaning that the acceleration will approach zero and hence velocity will approach a certain value, which in our universe is the speed of light, c.

That's the explanation in a nutshell. For more information look up special relativity.

- #8

DaveC426913

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It is not really a pointless question at all. There are motor designs that can do this in principle. Look up 'Bussard ramjets'.

Anyway, it will accelerate forever, yes, but as per Einstein's Special Theory of Relativity,

Here is a chart of acceleration vs. velocity over a fixed timeframe. Say your engine can accelerate your craft to 90% of c over the course of one year (a constant acceleration of a).

Code:

```
Time Elapsed Acceleration Velocity (as a per cent of c)
0 (Start) a 0
after one year a 90
after two years a 99
after three years a 99.9
after four years a 99.99
after five years a 99.999
...
```

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- #9

electricsheep

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but as per Einstein's Special Theory of Relativity,a constant acceleration doesn't result in a constant increase in velocity.

Hi DaveC, I may be nitpicking here, but when I studied special relativity back in high school, I had a lot of trouble with statements such as this one, because it doesn't specify the frame of reference from which you're viewing the body. I feel it's important to specify that here, "constant acceleration" is the "apparent acceleration" from the point of view of someone who is, say, riding on the engine.

This link explains it more clearly than I do. http://physics.nmt.edu/~raymond/classes/ph13xbook/node59.html" [Broken]

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- #10

DaveC426913

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Hi DaveC, I may be nitpicking here, but when I studied special relativity back in high school, I had a lot of trouble with statements such as this one, because it doesn't specify the frame of reference from which you're viewing the body. I feel it's important to specify that here, "constant acceleration" is the "apparent acceleration" from the point of view of someone who is, say, riding on the engine.

This link explains it more clearly than I do. http://physics.nmt.edu/~raymond/classes/ph13xbook/node59.html" [Broken]

Agreed. But I thought that a fairly simple question could be given a simple answer without overdoing it.

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- #11

Artman

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No. It could not have infinite top speed. Not even hypothetically. If it could, it would not move at all and be where it was going.

- #12

fedorfan

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Alright, I see what yall are saying here, that the faster you go the heavier you get. Oh, and I didnt mean infinite top speed that way, I meant it wouldn't stop accelerating. That bussard ramjet is very interesting and an ingenious but simple idea. Why haven't they made one yet?

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- #13

DaveC426913

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Lot of info out there about it - it's been studied. The technology is far beyond current ability.That bussard ramjet is very interesting and an ingenious but simple idea. Why haven't they made one yet?

- #14

fedorfan

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- #15

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This is an adequate explanation in the light of the question asked, but one should note that E=mc^2 refers to the rest energy of an object, which stays constant under acceleration by an external force.The faster something moves, the more kinetic energy it has. E=mc^2, so the faster it moves, the heavier it becomes...

It is really the relativistic momentum that increases with speed and approaches infinity in any inertial reference frame if the speed in that frame approaches the speed of light. The relativistic momentum is given by:

[tex]p = \frac{mv^2}{\sqrt{1-v^2/c^2}}[/tex]

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- #16

electricsheep

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This is an adequate explanation in the light of the question asked, but one should note that E=mc^2 refers to the rest energy of an object...

Actually, I don't think E is the rest energy, because that would imply m is the rest mass, which is not incorrect but entirely irrelevant to what I was explaining there. I'm fairly sure E is properly defined as the "mass energy", which is just the amount of energy equivalent to a certain mass. In our case of accelerating a body, this "mass energy" originates from an increase in kinetic energy. Everything I've (quickly) found online so far doesn't contradict what I said. Please provide me with a reference if I'm wrong.

It is really the relativistic momentum that increases with speed and approaches infinity in any inertial reference frame if the speed in that frame approaches the speed of light. The relativistic momentum is given by:

[tex]p = \frac{mv^2}{\sqrt{1-v^2/c^2}}[/tex]

You've lost me there. You seem to be saying that if the speed increases, then the relativistic momentum increases. I agree but that's nothing to do with our discussion. And your expression is also incorrect. It should be:

[tex]p = \frac{{m_0}v}{\sqrt{1-v^2/c^2}}[/tex]

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c1"

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- #17

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You are not wrong, just the modern concept of mass is different. SeeActually, I don't think E is the rest energy, because that would imply m is the rest mass, which is not incorrect but entirely irrelevant to what I was explaining there...

Please provide me with a reference if I'm wrong.

http://www.geocities.com/physics_world/mass_paper.pdf

If you search the relativity forum on this, you will also find lots of discussion on it.

This is the "older" definition. The use of the termsAnd your expression is also incorrect. It should be:

[tex]p = \frac{{m_0}v}{\sqrt{1-v^2/c^2}}[/tex]

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c1"

Also, the total relativistic energy is lately given by: [tex]E = \frac{mc^2}{\sqrt{1-v^2/c^2}}[/tex], where [tex]m[/tex] is the proper mass, or rest mass, or just mass...

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- #18

electricsheep

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You are not wrong, just the modern concept of mass is different. See

http://www.geocities.com/physics_world/mass_paper.pdf

Nice.

This is the "older" definition. The use of the termsrest massandrelativistics massis on the decline, but still found in plenty of writings, making it somewhat confusing sometimes.

That wasn't what I was trying to point out. Your original expression for relativistic momentum had [tex]v^2[/tex], which made me do a double take and do a quick search on the web just to confirm my memory. But yes, I think I'm thread-jacking now. My apologies.

- #19

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That wasn't what I was trying to point out. Your original expression for relativistic momentum had [tex]v^2[/tex], which made me do a double take and do a quick search on the web just to confirm my memory. But yes, I think I'm thread-jacking now. My apologies.

Oops! I'm sorry for having missed that [tex]v^2[/tex] - it's surely wrong in momentum!

I think to conclude this "diversion" (started by me), it is better to work with relativistic total energy that diverges to infinity as an object nears the speed of light than to work with relativistic mass that diverges. But, in the end, it is a matter of choice, I guess.

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