- #1

- 18

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- B
- Thread starter Zarich12
- Start date

- #1

- 18

- 0

- #2

mfb

Mentor

- 35,908

- 12,733

You can divide or multiply the equations by the speed of light or any other non-zero variables as often as you want. It does not matter which system of units you use, as long as you keep them consistent.

Energy divided by time is power, right.

- #3

- 18

- 0

- #4

Nugatory

Mentor

- 13,842

- 7,237

You take the mass times the speed of light (in miles per second?) and square that. Over one second that is equal to the number of watts the object could produce.

If you're measuring energy in joules and power in watts, you'll need to use meters for distances and meters per second for speeds.

- #5

- 18,485

- 8,387

- #6

Bandersnatch

Science Advisor

- 3,223

- 2,431

- #7

mfb

Mentor

- 35,908

- 12,733

The equation is solved for E already: it is "E=something".

- #8

jtbell

Mentor

- 15,866

- 4,451

In more explicit mathematical terms, ##mc^2## means ##m(c^2)## not ##(mc)^2##.

- #9

- 18

- 0

Right, sorry about that. I meant meters.If you're measuring energy in joules and power in watts, you'll need to use meters for distances and meters per second for speeds.

- #10

- 18

- 0

Just the speed of light.

- #11

- 18

- 0

Right, and given that you know only to square the speed of light, not the entire equation.In more explicit mathematical terms, ##mc^2## means ##m(c^2)## not ##(mc)^2##.

- #12

Ibix

Science Advisor

- 9,035

- 8,727

- #13

ChrisVer

Gold Member

- 3,381

- 464

then you have 1 equation with 2 unknowns as a result it does not have 1 solution. It's exactly the same as [itex]y= ax[/itex] if you know [itex]a[/itex]. If you know neither [itex]y,x[/itex] then this equation is solved by an infinite set of [itex]y,x[/itex] which lie on a line with slope [itex]a[/itex].Right, and given that you know only to square the speed of light, not the entire equation.

As things are you are only confusing yourself; what's the question you want to answer?

If you know the rest mass you input it in [itex]m[/itex] (in kg) and multiplying with the [itex]c^2[/itex] (in m/s) you can obtain the energy [itex]E[/itex] (in Joule) of that mass at rest. It's not an equation that needs a lot to be solved.

In order to speak for radiated away power, you must have an energy difference [itex]\Delta E = E_{fin}-E_{init}[/itex] in some time interval [itex]\Delta t = t_{fin} - t_{init}[/itex]:

[itex]Power= \frac{\Delta E}{\Delta t}[/itex]

The only way to have some [itex]\Delta E[/itex] is if your mass is changing with respect to time: ([itex]\Delta m= m_{fin} - m_{init} \ne 0[/itex].)

Last edited:

- #14

- 18

- 0

Share: