Using e=mc^2 to calculate electrical properties

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from rearranging e=mc^2 i got time equivalent to (kg m^2/(A V) (kilogram meter squared per ampere volt))^(1/3)

so if i have a material that is 1kg, 1m long, has 1amp, and 1volt, what does the time attribute mean? is that the amount of time it takes an electron to travel from one end of the object to the other?
 

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  • #2
Pengwuino
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Rearranging units randomly in this fashion is meaningless. Where do you even get the 't' from? the speed of light?
 
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  • #3
russ_watters
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That's not re-arranging equations, it is rearranging units! You can rearrange equations - you can't rearrange units.
 
  • #4
Pengwuino
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But Russ, I did say units..... *whistles innocently*
 
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_why_ can't you do that though? Obviously that cluster of units together has some sort of significance. For some reason 1 (kg m^2/(A V) (kilogram meter squared per ampere volt))^(1/3) is equal to 1 second. All I am wondering is why?
 
  • #6
ZapperZ
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_why_ can't you do that though? Obviously that cluster of units together has some sort of significance. For some reason 1 (kg m^2/(A V) (kilogram meter squared per ampere volt))^(1/3) is equal to 1 second. All I am wondering is why?
Because this is physics, not mathematics.

Each of those quantities have a physical significance. You can't simply manipulate those symbols without understanding the physics behind those manipulation. If you do that, you'll get absurd results.

For example, look at the units for torque, and for work done. If you simply put blinders on and forget about the physical meaning of each of those, and pay attention only to their dimensions, you'll think that they are the same thing. They are not.

Zz.
 
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Who said those were absurd results? that seems like its a perfectly normal result. I guess my question is, what _are_ the physics behind that equation?
 
  • #8
ZapperZ
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Who said those were absurd results? that seems like its a perfectly normal result.
The "time" factor in your original question is the "result".

I guess my question is, what _are_ the physics behind that equation?
Maybe you should have asked that first, and then, after understanding that, go on to the next step. Without that first step, you risk doing something based on either faulty or incomplete knowledge.

E=mc^2 has been discussed ad nauseum on here (look in either the Quantum Physics or the Relativity forums), and in many reputable websites. Try starting there first and see if there is anything you do not understand that we can try to clarify on here. That's the best way to learn something new, which is to try and understand it yourself first and then get some help in figuring out what you find puzzling or confusing.

Zz.
 
  • #9
russ_watters
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Who said those were absurd results?
Everyone who has responded so far!
that seems like its a perfectly normal result.
If it were a normal result, then the question would not need to be asked.
I guess my question is, what _are_ the physics behind that equation?
I'm sorry, but as already said, there are none.
 
  • #10
Pengwuino
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Yes, you have to know what the equation means. Think about a blind treatment in this fashion. The energy an object has is [tex]E = \frac{mv^2}{2}[/tex]. Oh but the energy is also [tex]E = mc^2[/tex]. So does [tex] c^2 = \frac{v^2}{2}[/tex] mean anything?
 
  • #11
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does it? I mean they have to have _some_ sort of significance together otherwise we would never end up in that combination.
 

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